Chapter 4 Chemical Kinetics

The role of a catalyst is to provide an alternative path by forming an activated complex with lower activation energy without changing the enthalpy of reaction.

Activation energy is required to start the reaction by making an activated complex.

A surface is provided by a catalyst for reactants so that activated complex can be achieved with lower activation energy.

Only option (iii) talks about activation energy hence it is correct.

No change in Gibbs free energy, so option (i) is wrong

No change in enthalpy of reaction, hence option (ii) is wrong.

The equilibrium constant remains the same with or without the use of catalyst, hence option (iv) is also wrong.

In the presence of a catalyst the Gibbs free energy does not change as the energy difference between reactant and product remains the same.

A reference can be taken from the above graph that final product AB energy and the reactants (A and B) energies are the same in the absence and the presence of catalyst remains the same in the presence or absence of a catalyst.

Thus option (iii) is correct.

Taking ln on both sides give: ln k= -E_A/RT + lnA

The above equation has E_A as activation energy and k as the rate constant.

The equation can be changed to a line equation i.e. y=mx+c, by taking log on both sides.

For two values of k at different temperatures we can get the value of E_A from above Equation.

Thus option (ii) is correct.

Other options do not give an equation or values for a graph that can be plotted to find the value E_A.

Activation energy is the energy required to reach the activated complex and for a forward reaction from the graph given above the sum of energy E₁ and E₂ is required to carry the reaction forward.

According to Gibbs free energy, the energy of the product should be less than reactant in order to be stable but in the following graph, the energy of the product is more than that of reactants energy, hence it is less stable than reactant.

Thus option (i) is correct.

Given,

For first-order reaction for a gaseous reaction at a constant temperature the concentration is replaced by the partial pressure of reactants and products.

Where p_a is the partial pressure of element A.

p_i is the initial pressure of element A.

Let p_t be the total pressure , p_t =p_a + p_b + p_c

Where, p_a ,p_b ,p_c are partial pressure of A,B,C respectively

At t=0 p_a=p_i , p_b =0, p_c =0

At t=t p_a=p_i-x, p_b=x , p_c=x ,so p_t= p_i-x + x +x p_t=p_i + x ………(p_t=p_a+p_b+p_c)

Know at t=t , p_t = p_i+x x=p_t-p_i

At t =t , p_a=p_i-(p_t-p_i) p_a=2p_i-p_t

Putting these values in Equation 1 we get,

K=

Thus Option (ii) is correct.

Given:

Taking ln on both sides we get

ln k = -+ lnA ………. Equation 1

As know following Equation 1 represents the equation of line i.e. y=mx+c,

Where m is the slope of the graph and c is y-intercept on comparing with Equation 1 we get

y is ln k, x is 1/T, m is -E_a/R, c is ln A

since the slope is negative and there is a y-intercept only one graph follows that graph 1.

In graph 2 the slope is positive so it is incorrect.

In graph 3 the slope is positive so it is incorrect.

In graph 4 the slope is positive so it is incorrect.

The given Arrhenius equation can be written as

Here, A is the Arrhenius constant, E_a is the activation energy, R is Universal gas constant and T is temperature.

If we differentiate both sides with respect to time (t) we get,

……… Equation1

Differentiation of A and R with respect to time t is zero as they are constant values.

As we know,

……. Equation2

Rate constant increases exponentially with time when E_a decrease and T increases with the passage of time.

As shown in equation 1 and equation 2.

Thus option (iv) is correct.

The graph is a straight line till V₃ or at t=40 we know the equation of a line.

y=mx + c, where m is the slope which can be determined by using the equation.

where (x₁,y₁),(x₂,y₂) are point on line,

For the above graph x₁=0,x₂=40,y₁=0 and y₂=V₃ .

Putting these values in the above equation we get slope V₃/40.

The average rate is equal to the slope of the graph hence V₃/40.

Thus option (iii) is correct.

for a general equation Aa + bB cC + dD

Let Rate =k[A]^x[B]^y

Order of a reaction is the sum of the power of molar concentration of A and B reactant in the rate law equation (x and y) making a statement (iv) correct.

Order can be 0,1,2,3 or even fractional number depending upon x and y making a statement (i) correct.

x and y can be or cannot be equal to the stoichiometric coefficients of reactants of a balanced chemical reaction hence the order are determined experimentally making the statement (iii) false and statement (ii) true.

The slope of a line m=(y₂-y₁)/(x₂-x₁) ,where (x₁,y₁),(x₂,y₂) are points through which graph passes.

Average rate of reaction at 40 seconds can be determined by determining the slope of line which can be evaluated from the above equation option ii,iii, iv (as the graph passes through these points (50, V₄),(40, V₃),(30, V₂),(20, V₁) respectively) follow the above equation and give the correct and same slope of line.

The Option (i) is false as there is no such (50, V₅) point through which the graph passes.

The rate of a reaction can be defined as the change in concentration of a reactant or product in unit time.

The rate of a reaction decreases with the passage of time.

With the continuation of the reaction, the concentration of a reactant decreases with time.

The rate of a reaction is directly proportional to the concentration of reactant so, if the concentration of a reactant is decreasing, then the rate of reaction will also decrease with time.

As an example take equation

(NEGATIVE SIGN SINCE THE RATE IS GETTING NEGATIVE AS REACTANTS ARE DECREASING ON PASSAGE OF TIME).

Rate of a reaction is defined as the change in the concentration of the reactant or product in unit time.

Rate= =

Where R is the change in concentration of a reactant in a time interval t and P is the change in concentration of a product in a time interval t.

For such type of a reaction, the rate is given by dividing the rate of appearance or disappearance of product and reactant respectively by their stoichiometric coefficients hence the rate of reaction become,

Rate =

Rate=

Thus option (iii) is correct.

For an exothermic reaction the energy of reactants should be greater than that of the product.

For graph (a) follows the statement, and when the reaction goes in the forward direction the difference in energy of reactant and product is released in the form of heat so the reaction is an exothermic reaction.

For Graph (b) the energy of the product is greater than that of a reactant. Hence, the reaction is endothermic in nature for the forward direction.

In graph (c) the energy of product and reactant is same. Hence, it is neither endothermic nor exothermic in nature.

Given,

Rate_O = k[A][B], the rate is directly proportional to the unit power of the concentration of element B. So, if its concentration is doubled the rate of reaction is also doubled.

Rate=k[A][2B] 2K[A][B] 2Rate₀

Thus Option (ii) is correct.

collision theory states that

Rate= , Z_AB is the effective number of the collision of reacting molecules and atoms considered to be hard-sphere and the exponential term is the energy of these molecules above E_a for complex molecules all the collision does not lead to the formation of the product hence proper orientation is required for the collision.

Given, t_1/2=1.2610¹⁴ we know the reaction is a first order

t_1/2=.693/k ……..for first-order reaction

using above given value we can find k

For 100% completion

At 100% completion R=0 so

Thus Option (iv) is correct.

We know that the

Rate=k[A]^x[B]^y

To determine the value of x and y, We will use the data given in the table

By observing that when the concentration of B is doubled keeping the molar concentration of A constant the rate of production of C changes from .10 to .40, it means then the rate of production of C or Rate is directly proportional to the square of the molar concentration of B.

By keeping the concentration of B constant and on doubling the concentration of A from .30 to .60, the rate of production of C changes from .1 to .2, this means that the rate of production of C or Rate is directly proportional to the molar concentration of A.

Hence, x=1 and y=2,

Rate=k[A][B]²

The value of x and y, for the other options, do not give the correct value rate of production of C for the respective value of A and B.

The catalyst does not alter the DG (change in Gibbs free energy also called Delta G ()), as the energy of reactant and product remains the same in the presence or absence of a catalyst so, the net change or difference of their energy remains the same.

Thus making option (ii) incorrect.

One can observe from the graph given below that a catalyst catalyzes the forward and the backward reaction to the same extent.

Thus making option (i) correct.

Catalyst provides an alternative path with the lower activation energy for a reaction and does not change the equilibrium constant of a reaction.

Thus making option (iii) and option (iv) correct.

Conditions play an important role in determining the order of a reaction.

For Example, Hydrolysis of .01 mole of ethyl acetate in the presence of 10 moles of water, the reaction at first glance appears to be the first-order reaction but since the concentration of water does not change much in comparison to ethyl acetate the order of reaction becomes pseudo-first-order reaction.

For a pseudo-first-order reaction, the reactant with a large amount of concentration is ignored.

As the change in its concentration is negligible with comparison to another component in respect with time.

Hence, the rate only gets dependent upon reactant in small quantity.

Thus making option (i) correct and other options are incorrect.

For a given reaction the concentration of A and B varies exponentially with time as given when the reaction will begin the concentration of A will be maximum and with the passage of time will reduce and moves towards zero with exponentially i.e. becoming asymptotic to the x-axis.

Similarly, the concentration of product B will increase exponentially with the passage of time and will become constant. Towards the completion of the reaction the concentration of product B will increase with a small rate as concentration of A on L.H.S has reduced significantly.

IMPORTANT FACT: L.H.S means Left hand side of reaction.

Rate law cannot be determined if a reaction is reverse as the product formation and dissociation takes place simultaneously similar for its reactant hence impossible to plot a graph of concentration vs time.

Rate law cannot be determined for a reaction involved in steps as different intermediate and by-products are simultaneously formed in those step hence difficult to monitor their concentration with time.

Rate law can not be determined for a reaction in which reactant is in excess since its rate of disappearance will be constant at every time.

For an elementary reaction (Reaction in which one or reactants react in a single step to for a product) the molecularity and the order is the same for example

A + BProducts (single-step elementary reaction)

R=k[A][B]

Order = 1+1=2

Molecularity is 2, therefore, molecularity and order are the same.

Molecularity is the number of the reacting species (atoms or molecules) that must collide in order to give a stable product.

Hence the molecularity of a reaction can never be zero or non-integer for a complete reaction.

Since there is only one reacting species in a unimolecular reaction. Thus it is involved in the rate-determining step.

For a unimolecular reaction, molecularity is one as only one reacting species are involved in the reaction.

Thus option(i) is correct.

Since the rate of the reaction depends upon the molar concentration of one reactant, the order of the reaction is also one.

Thus option (ii) is also correct

Option (iii) is false as the order of a reaction is 1, not 0.

A ……….. (unimolecular reaction)

Rate=k[A]

Order= 1

Molecularity= 1

Thus option (iv) is correct.

The molecularity is the number of the reacting species (atoms, molecules) in a reaction occurring in a single step. These are the species that collide in order to form a product.

Hence, the molecularity of a complete reaction can never be zero or non-integer.

For the complex reaction molecularity of the slowest step of the reaction is always equal to the order of the overall reaction.

As the number of reacting species in the slowest step is equal to the experimental value of the order of the overall reaction.

for a zero-order reaction

R=k[A]⁰R=k where R is the rate of reaction and k is Rate constant and A is the molar concentration of reactant,

Thus Statement (i) is true.

Thus Statement (ii) is false, As Rate does not depend upon the concentration of ammonia.

Since the rate of decomposition of ammonia is equal to the rate constant (a constant value) the decomposition of ammonia will continue until ammonia does not vanish completely.

Thus making option (iii) correct.

In the presence of a catalyst at high-pressure, gas molecules of ammonia saturate the surface of the catalyst thus a further change in pressure does not alter the amount of ammonia on the surface of the catalyst, hence no effect of the increase in pressure.

Thus making option (iv) incorrect.

The activated complex which is formed by the absorbing of activation energy, so its energy becomes more than that of product and reactants.

Being highly unstable (due to high enthalpy) it breaks down into stable product or products, releasing out energy making Option (i) correct.

The Gibbs free energy of the reaction becomes less than zero (<0) due to the evolution of energy on decomposition, so the stability of product or products increases.

Since all the activated complex does not form product, some of it also changes to reactant as the energy of the reactants are also less than that of the activated complex making option (iv) correct.

The Maxwell Boltzmann distribution of energy can be further studied from the graph below,

Which shows a decrease in most probable Kinetic Energy (peak of the graph) due to the increase in collisions among the molecules of gas with an increase in temperature making option (iv) correct.

There is also an increase in the fraction of molecules having energy greater than E_A and most probable Kinetic Energy due to the increase in temperature T, as the molecules of gas get thermally excited due to the increase in temperature making option (ii) correct.

Maxwell Boltzmann distribution of energy.

The area under the graph remains constant since the total energy of all molecules present in a sample remains 1 for all time even on increasing temperature making a statement (i) correct.

The most probable energy decrease due to the collision between molecules but the number of molecules with activation energy E_A increases due to the increase in temperature.

The temperature curve shifts right and get broadens at higher temperature due to the involvement of additional molecules whose energy goes above activation energy due to an increase in temperature and decrease in the most probable energy of the molecule making a statement (iv) also correct.

Arrhenius equation follows k=A

Rate is directly proportional to the rate constant

The rate of reaction increase with the increase in the rate constant and the rate constant increases with increase in temperature as the exponential value () decreases and the numerator (A) remains constant so, the L.H.S value increases hence option (i) is correct.

A decrease in value of activation energy increases exponential value so rate constant increases as the rate are directly proportional to the rate constant its own value also increases simultaneously hence option (ii) is correct.

Option (iii) is incorrect as of the rate constant value increases exponentially with an increase in temperature as the denominator term () decreases with increase in temperature so L.H.S value increases as the numerator is a constant.

Option (iv) is incorrect as the rate of reaction increases with the decrease in activation energy value as clarified in option (ii).

The important fact the rate of reaction is proportion directly to rate constant k.

the statement (ii) is incorrect since the catalyst provides an alternative path of reaction with the lower activation energy, it itself does not change the activation energy.

statement (i) is correct since the function of a catalyst is to provide an alternative path for the reaction mechanism for lower activation energy.

Statement (iv) is incorrect as there is no change in the Gibbs free energy of the reaction since no alteration is done by the catalyst on the energy of reactant and product.

statement (iii) is incorrect as the catalyst provide the alternative path by forming an activated complex with lower energy it itself does not raise or lowers the activation energy of a reaction.

for a zero-order reaction,

Rate=k[R]⁰

R is the molar concentration of reactants.

Hence the reaction rate remains constant with time and becomes zero when the concentration of reactants is zero.

At t=0, R=R₀ and at t=t, R=R, where R is the concentration of reactants at time t.

Integrating w.r.t. t on both side and putting limit we get,

[R]=-kt + [R₀]

Following an equation of a line with negative slope, hence the graph (iv) is correct.

For a first order reaction the Rate = k[R] solving it

At t=0 value of R =R₀ and at t=t value of R=R …(let),

On integrating w.r.t t on both sides we get,

……………………. Equation 1

Here, R is the molar concentration of a reactant at time t.

Taking ln on both the side we get,

………………… Equation 2

ln =2.303log

for t_1/2 at t=t_1/2 [R]=[R₀] for first-order reaction

so k=

t_1/2 =(2.303 log2)/k t_1/2 is independent of [R₀]………. Equation 3

t_1/2=.693/k

so from Equation 1,2,3, the above graph can be plotted t_1/2 independent of [R₀] so the line is constant and molar concentration is exponentially related with time.

The Equation 2 gives an equation of a line i.e. y=mx +c, with m=k/2.303 and c=0.

A condition under which the order of a chemical reaction can be altered is by taking the solvent or the reactant in the excess amount due to which its concentration does not change much.

For example,

CH₃COOCH₃ + H₂0 ⇌ CH₃COOH + CH₃OH

Initially, rate is given by,

Rate= k’ [CH₃COOH] [H₂O]

When we add an excess of solvent then,

Rate= k[CH₃COOCH₃]

Here k= k’[H₂O]

This is also known as pseudo first-order reaction.

For the order of the reaction to be zero the powers of the concentration of the reactants must be equal to zero. In the zero order reactions the rate of the equation is equal to the rate constant.

Rate equation for the reaction 2A + B → C

Rate = k[A]⁰[B]⁰

2NO (g) + O₂ (g) → 2NO₂ (g)

The possible mechanism for the reaction will be

i) NO + O₂ ⇌ NO₃ (fast)…………………k(equilibrium constant)

ii) NO₃+ NO⇌ NO₂ + NO₂ (slow)…………………..k’

Rate is determined by the slow step, so

Rate = k’[NO₃][NO]…………………1)

K= [NO₃]/[NO] [O₂]

K[NO] [O₂] = [NO₃]………………a)

By putting values of a) in 1)

Rate = k’k[NO][O₂][NO]

=k’k[NO]²[O₂]

The rate of the reaction can be determined by

Rate = k[NO]²[O₂]¹

For an elementary reaction, order is same as molecularity. Elementary reactions are those reactions which occur in a single step.

Suppose there are 2 reactants A and B which form a product C. Then

A+B → C

Rate is determined by,

Rate = k[A]¹[B]¹

Order of the reaction is 2.

Molecularity of the reaction is also 2 as there are total 2 molecules of reactants that are reacting for the formation of a product C.

Suppose reactant A forms the product B. Then the reaction can be written as

A → B

Rate of the reaction is given by

Rate₁= k[A]^x …………………..i)

x is equal to the order of the reaction.

When A is tripled rate of reaction becomes 27 times that is,

Rate₂ = k [3A]^x = 27 Rate₁ ………………………ii)

Dividing equation i) and ii)

Rate₁/ 27Rate₁ = k [A]^x/ k[3A]^x

1/27 = [1/3]^x

x = 3

Order of the reaction is 3.

Consider a reaction,

R → P t=0

Here R is the reactant and P is the product.

Rate = k[R]⁰ t=t

(Instantaneous rate) –dR/dt = k

dR =-kdt

On integrating both sides,

[R] = -kt + I

At t = 0,

[R]= [R₀] which makes I= R₀

[R] = [R₀]– kt. ………………………..i)

Here [R] = concentration of reactant at time ‘t’.

[R₀] = initial concentration of the reactant.

This reaction is known as the integrated rate equation of zero order.

After completion of zero order reaction [R]= 0

Using [R]= 0 in i)

[R₀]=kt

t=[R₀]/k

No, the reaction is not an elementary reaction. As in case of an elementary reaction molecularity and order are same. Molecularity is the number of atoms colliding to react. So if it would have been an elementary reaction then the order of B with respect to A had been 1 instead of 3/2.

The two most important conditions for a reaction to occur according to the collision theory are:

• Energy greater than activation energy.

• Proper orientation of reactant molecules at the time of collision.

The equation of the collision theory is given by

Rate = Pz_ABe^-Ea/RT

So when the reaction occurs then molecules that don’t have a proper orientation decreases the rate of a reaction.

No, the molecularity of a reaction is never zero. Molecularity of any reaction is the number of reacting species taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction. So, zero molecularity means there is no reactant, as a result of which a reaction cannot occur. So if there is a reaction then its molecularity will be greater than zero.

(i) It is a zero order reaction as the graph is satisfying the equation [A] = [A₀] – kt.

(ii) The slope of the curve is the negative of the rate constant that is denoted by –k.

(iii) Unit of rate constant is Ms^-1 or mol L^-1s^-1.

In order to form a product it is important that the colliding species must possess an energy greater than the activation energy. So, at room temperature the energy is less than the activation energy. According to Maxwell Boltzmann energy distribution curve when temperature T becomes T+10 degree Celsius then the effective collision and energy of molecules increases resulting in the formation of product.

Rate of reaction increases with the rise in temperature because according to the Maxwell-Boltzmann Energy distribution curve as temperature increases, fraction of molecules having energy greater than activation energy also increases, which results in the increase in number of effective collisions and thus rate constant and rate of reaction increases.

Oxygen is available in plenty of air yet fuels do not burn by themselves at room temperature because reactants must have a minimum amount of energy known as the activation energy in order to form a product. At room temperature, molecules don’t have required amount of energy. This energy is also less than the activation energy. When we increase the temperature then the energy becomes greater than the activation energy.

The probability of reaction with molecularity higher than three is very rare as molecularity defines the collision of reactant molecules due to which the reaction occurs. The two most important conditions for a reaction to occur are:

• Energy greater than activation energy.

• Proper orientation of reactant molecules at the time of collision

So, for a reaction having more than three molecules, the proper orientation is not possible that makes these reactions rare.

Rate = Pz_ABe^-Ea/RT

P= probability factor.

The rate of a reaction generally decreases with the course of the reaction as the concentration of reactant decreases as soon as the formation of product starts. So during the course of a reaction the concentration of reactant becomes less than that present initially. As rate is directly proportional to the concentration, therefore the rate of reaction also decreases with time.

Thermodynamic feasibility of the reaction alone cannot decide the rate of the reaction because for a reaction to occur it is important for the molecules to have an energy greater than the activation energy. So in order to increase the energy usually we increase the temperature so that the rate of the reaction becomes faster.

For example,

Diamond → Graphite ΔG=-ive

ΔG negative means that the reaction is feasible but this reaction is a slow process as energy is less than the activation energy.

We heat oxalic acid solution because without heating it is a slow process as an energy greater than the activation energy is required for a reaction. So in order to increase the energy, temperature must be increased which is only possible by heating the oxalic acid solution. So, when the temperature increases molecules having energy greater than activation energy also increases which increases the rate constant of a reaction.

Molecularity of any reaction cannot be equal to zero as molecularity is the number of reacting species taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction. So, zero molecularity means there is no reactant, as a result of which a reaction cannot occur. So if there is a reaction then its molecularity will be greater than zero.

Molecularity is determined theoretically whereas order is a practical quantity. Molecularity is applicable only for elementary reactions as they are the single step reactions and the rate depends on the concentration of each molecule, whereas in case of complex reactions there are multiple reactions involved and thus molecularity holds no meaning. Molecularity is the collisions of reactant molecules and to increase the rate it requires a proper orientation at the time of collisions which is only possible in less molecules.In complex reactions order is determined by the slow step and it can also be easily determined in the elementary reactions as well.

We cannot determine order of a reaction by taking into consideration the balanced chemical equation as order is an experimental quantity. There are some reactions in which not all the molecules are used to determine the rate or order of the reactions. For example, in complex reactions the rate and order is dependent on the slow step reactions, so in such cases the order is not completely dependent on the balanced chemical equations.

(i)→(a),(ii)→(b),(iii)→(b),(iv)→(a)

(i) →(a),(iv) →(a)

For the first order of a reaction, the rate is directly proportional to the molar concentration of the reactant.

Thus,

Rate=-d[R]/dt=kR⟹-(d[R])/R=kdt

Thus, The above equation gives graph (i) for a first-order reaction.

Integrating both sides from t=0 to t=t at concentration R=R0 to R=R respectively.

We get,

log⁡〖 [R]〗=-kt/2.303+log⁡[R_0]

Where K is the rate constant, R is the molar concentration of the reactant at time t and R0 is the molar concentration of a reactant at time t=0.

Above equation represents the equation of a line i.e. y=mx+c, for y=log [R] and x=t.

Slope is -k/2.303.

Thus, (iv) →(a).

(ii) →(b),(iii) →(b)

For a zero-order reaction, Rate is only depending on rate constant.

Rate=k〖[R]〗^0

Where k is the rate constant and R is the molar concentration of a reactant.

Thus, the rate is constant with respect to the concentration of a reactant. Giving, graph (ii) for a zero-order reaction.

From the above equation, we get,

-(d[R])/dt=k

Integrating both sides from t=0 to t=t at concentration R0 to R respectively. We get,

[R]=-kt+[R_0]

Where R is the molar concentration of a reactant at t=t, R0 is the molar concentration of a reactant at t=0.

The above equation represents the equation of a line i.e. y=mx+c.

Thus, for a concentration vs time graph for a zero-order reaction will be a straight line with slope equals -k.

Thus, the graph (iii) is for a zero-order reaction.

(i)→(c),(ii) → (a),(iii) → (d),(iv) → (f),(v) → (b),(vi) → (e)

(i) →(c), A catalyst provides an alternate path of reaction with a lower activation energy between the reactant and the product. Thus, lowering the potential energy barrier.

(ii)→(a), Molecularity of a reaction is the number of reacting species (atoms, molecules) taking part in an elementary reaction that must collide in order to bring about a chemical reaction.

Thus, for an elementary reaction taking place molecularity cannot be zero or fraction for it.

(iii) → (d), the first half-life of a first-order reaction is, t_(1/2)=.693/k.

For the second half-life of the first-order reaction, the initial concentration of reactant after the first half-life is R_0/2 , the final concentration of reactant after the second half-life is R_0/4 .

Thus, putting these values in the equation

log⁡〖 [R]〗=-kt/2.303+log⁡[R_0]

Where K is the rate constant, R is the final molar concentration of a reactant and R0 is the molar concentration of a reactant initially.

We get,

t_(1/2)=0.693/k

Hence, the second half-life of the first-order reaction is equal to the first half-life of the first-order reaction.

(iv) →(f), An Arrhenius equation i.e. k=Ae^(-E_a/RT) .

Where A is the frequency factor and term e^(-E_a/RT) corresponds to the fractions of molecules having kinetic energy equal to or greater than the activation energy.

(v) →(b), An Arrhenius equation i.e. k=Ae^(-E_a/RT) .

Where A is the frequency factor and term e^(-E_a/RT) corresponds to the fractions of molecules having kinetic energy equal to or greater than the activation energy.

Thus even if the reaction is an energetically favorable reaction but if there is not the proper orientation between the collision of the molecules the reaction will be slow.

(vi) →(e), Area under the Maxwell Boltzman curve is constant as the total probability is one.

The above statement means that the sum energies of all molecules involved in a reaction remain constant even on increasing the temperature.

The most probable energy of molecules decreases due to the increased in the collision between molecules but the number of molecules with activation energy EA increases due to the increase in temperature.

Thus, the total energy remains constant.

(i)→(b),(ii)→(a),(iii)→(c)

(i) →(b) Thermodynamics states that the diamond will convert to graphite. Since the Gibbs free energy of graphite is less than that of the diamond.

The rate of conversion of diamond to graphite is so slow that the change is not perceptible at all.

(ii) →(a) Instantaneous means for an instant or short period of time.

Thus, the instantaneous rate is the slope concentration of reactant vs time graph i.e. - (d[R])/dt.

(iii) →(c) The average rate of a reaction is the rate of appearance or disappearance of product or reactant respectively.

The time interval is large to get the large value of change in concentration of product or a reactant.

〖Rate〗_avg=-∆R/∆T=∆P/∆T

Where ∆R is the change in the molar concentration of a reactant, ∆P is the change in the molar concentration of product and ∆T is the time interval.

(i)→ (b),(ii)→ (a),(iii)→ (d),(iv)→ (c)

(i) →(b),

Rate of a reaction is the rate of disappearance of a reactant or appearance of a product.

Rate=-d[R]/dt=(d[P])/dt

Where R is the molar concentration of reactant and P is the molar concentration of the product.

The equation is known as Rate Law.

(ii) →(a), For a zero-order reaction, Rate is only depending on rate constant i.e. it is independent of the concentration of the reactant.

Rate=k〖[R]〗^0

Where k is the rate constant and R is the molar concentration of a reactant.

(iii) →(d), For a zero-order reaction, Rate is only depending on rate constant i.e. it is independent of the concentration of the reactant.

Rate=k〖[R]〗^0

Where k is the rate constant and R is the molar concentration of a reactant.

Thus, the above equation changes to,

Rate=k

Hence, the units of the rate constant are the same as that of the rate.

(iv) →(c), Complex reactions occur in the steps and the slowest step is said to be the rate-determining step as it involves the formation of the intermediate.

Thus, the order of the slowest step of the complex reaction is the order of the overall reaction.

Order of a reaction is an experimental quantity. Thus, it can be zero or fractional.

We cannot determine order from a balanced chemical reaction as a chemical reaction hardly completes in a single step.

Thus, both assertion and reason are correct but the reason is not the correct explanation of assertion.

Order of a reaction is an experimental quantity and it is the sum of the powers of the molar concentration of reactants in a rate law expression.

Whereas molecularity of a reaction is the number of reacting species that must collide in order to bring about a chemical reaction. For reaction taking place in steps molecularity is determined by the sum of stoichiometric coefficient the slowest step (which is the rate-determining step).

Thus, molecularity and order are different.

So, the assertion is wrong and the reason is correct.

A catalyst alters the path of a reaction by providing an alternative path with lower activation energy but there is no change in the enthalpy of reactants and products.

Catalyst forms an intermediate with lower activation energy thus lowers down the potential barrier of the reaction as shown in the diagram below.

Thus, both the assertion and the reason are correct and the reason is the correct explanation of assertion.

For a compound formation the molecules should have energy greater than e–Ea/R T and there should be a proper orientation of collision for better energy transfer and finally for the formation of a compound.

Collision without proper orientation or sufficient kinetic energy does not lead to the formation of a compound.

Thus, the assertion is false and the reason is correct.

An Arrhenius equation i.e. k=Ae^(-E_a/RT) .

Where A is the frequency factor and term e^(-E_a/RT) corresponds to the fractions of molecules having kinetic energy equal to or greater than the activation energy.

The rate constant can be determined for a reaction from the above equation if the different values of temperature are given and it is accurate for simple as well as complex reactions.

Reactant molecules with sufficient kinetic energy and with proper orientation of collision undergo a chemical change and lead to the formation of a product that is why frequency factor is added in Arrhenius Equation.

Thus, the assertion is true and the reason is false.

The two most important conditions for a reaction to occur according to the collision theory are:

• Energy greater than activation energy.

• Proper orientation of reactant molecules at the time of collision.

The equation of the collision theory is given by

Rate = PzABe-Ea/RT

So when the molecules are not oriented properly then then no chemical change occurs.

For example,

(Proper orientation)

The most probable Kinetic energy and energy of activation are affected with increase in temperature as according to Maxwell Boltzmann energy distribution curve when temperature T becomes T+10 degree Celsius then the effective collision and energy of molecules increases resulting in the formation of product. Rate of reaction also increases with the rise in temperature because according to the Maxwell-Boltzmann Energy distribution curve as temperature increases, fraction of molecules having energy greater than activation energy also increases, which results in the increase in number of effective collisions and thus rate constant and rate of reaction increases.

When a catalyst is added to a reaction then according to the intermediate complex theory it provides an alternative path to the reaction or the reaction mechanism by reducing its activation energy between the reactants and products and hence lowers the potential energy barrier. Consequently, the rate constant of the reaction increases thus increasing the rate of the reaction. but all this does not affect the enthalpy of the reaction as catalyst does not alter the Gibbs energy change or extent of the reaction.

For a reaction, A+B → C, graph will be,

As the initial and final point of the reaction is same therefore enthalpy of the reaction is constant.

Without catalyst,

lnK1 = lnA – Ea1/RT ……………………………..i)

lnK2 = lnA – Ea2/RT……………………………….ii)

by substituting value of lnA in eq i) from eqii)we get,

lnK2 – lnK1 = 1/RT(Ea1-Ea2)

log K2/K1 = (Ea1-Ea2)/2.303RT

Difference between instantaneous rate of a reaction and average rate of a reaction.

Average rate of a reaction Instantaneous rate

The rate of change of concentration of species for a finite interval of time. Instantaneous rate is the rate of a reaction at a specified or particular point of time.

It is meant for a time interval. Instantaneous rate is for a point of time in a reaction when dt →0.

It is given by,

Ravg= -1/a(Δ[R]/Δt) = 1/b(Δ[R]/Δt It is given by,

Rinst = -d[R]/dt = +d[P]/dt

At a point dt →0.

Pseudo first order is a condition under which the order of a chemical reaction can be altered by taking the solvent or the reactant in excess amount due to which its concentration does not change much.

For example,

CH3COOCH3 + H20 ⇌ CH3COOH + CH3OH

Initially rate is given by,

Rate= k’ [CH3COOH] [H2O]

When we add excess of solvent then,

Rate= k[CH3COOCH3]

Here k= k’[H2O]

This is also known as pseudo first order reaction.