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Option : (d) zero

Option : (a) 0.05 s^-1

Option : (d) k = \(\frac{2.303}{t}\)\(log_{10}\frac{a}{(a-x)}\) 

Option : (b) one

Option : (d) mol L^-1s^-1

Because temperature and activation energy are inversely proportional to each other.

(iii) Collision of atoms or molecules possessing sufficient threshold energy results into the product formation.

Rate = [A][B]²

(i) New rate = 9 Rate 

(ii) New Rate = 8 rate

(i) Both assertion and reason are correct and the reason is correct explanation of assertion.

(i) Both assertion and reason are correct and the reason is correct explanation of assertion. 

Atm^-1/2sec^-1 

Activation energy : 

The energy required to form activated complex or transition state from the reactant molecules is called activation energy.

OR

The height of energy barrier in the energy profile diagram is called activation energy.

(iii) Assertion is correct but reason is incorrect. 

The collisions in which the molecules collide with sufficient kinetic energy (threshold energy) and proper orientation so as to result in chemical reaction are called effective collisions.

(iii) Assertion is correct but reason is incorrect.

The minimum energy required for the reactants to form activated complex is called Activation energy.

the minimum quantity of energy which the reacting species must possess in order to undergo a specified reaction.

Activation energy of a reaction decreases in presence of catalyst.

The collision that leads to the formation of product is called effective collision and the no. of collisions per second per unit volume is called collision frequency.

Rate constant has different units for different order of reaction but rate of reaction has fixed units. 

K.E of maximum fraction of molecules is called most probable K.E.

From an examination of above data, it is clear that when the concentration of B₂ is doubled, the rate is doubled. Hence the order of reaction with respect to B₂ is one. Further when concentration of A is doubled, the rate remains unaltered. So, order of reaction with respect to A is zero. 

                    OR

The probable rate law for the reaction will be

dx/xt = k[B₂]¹ [A]⁰ = k[B₂]¹

 Alternatively Rate = k[B₂]^x

1.6 × 10^-4 = k[0.5]^x

3.2 × 10^-4 = k[1]^x

On dividing we get x = 1

Rate = k[A]⁰ [B₂]¹ = k[B₂]

Threshold Energy: It is the minimum amount of energy which the reactant molecules must possess for the effective collision in forming the products.  

Activation Energy: It is the excess energy required by the reactants to undergo chemical reaction. 

Activation energy = Threshold energy – Average kinetic energy of molecules. 

Apart from the energy considerations, the colliding molecules should also have proper orientation for effective collision. This condition might not be getting fulfilled in the reaction.

The energy requied by the reacting species to form the activated complex is called Activation energy.

Minimum energy associated with reacting species to cross over the potential barrier is called threshold energy.

E_th = Average K.E + E_a

 Increased rate =( Increased conc.)ⁿ

=3² 

 =9

Rate of formation of Y increases by 9 times 

Integrated rate equations

The reaction follows second order kinetics. Therefore, the rate equation for this reaction will be:

If the concentration of X is increased to three times, then the rate of formation will increase by 9 times.

K = Ae ^-Ea/RT A=Frequency factor, Ea = Activation Energy

(a) Addition of the two steps gives the overall reaction

2O_3(g) ------> 3O_2(g)

(b) Intermediate is O_(g).

(c) The first step is unimolecular. The second step is bimolecular.

(a) N₂O is the reaction intermediate since it is formed in the first step and removed in the second step.

(b) By rate law, 

Rate = k [H₂][NO]². 

Since the first step involves the substances H₂ and NO, it is the slow and rate-determining step.

Given,

2H_2(g) + 2NO_(g) → 2H₂O_(g) + N_2(g)

Order of reaction in H₂ = \(n_{H_1}\) = 1

Order of reaction in NO = n_NO = 2

Rate constant = k = 0.42 M^-2s^-1

[H₂] = 0.015 M; [NO] = 0.025 M

Rate of reaction = R = ? 

By rate law,

Rate = R = k [H₂] [NO]²

= 0.42 x 0.015 x (0.025)² M^-2s^-1

= 3.94 x 10^-6 Ms^-1

∴ Rate of reaction = R = 3.94 x 10^-6 Ms^-1

Option : (b) a straight line with a positive slope.

Option : (c) \(\frac{1}{2k}\)

The time taken for the concentration to change from 0.1 M to 0.25 M is 60 minute.

(d) Energy of activation as well as the frequency factor.

1. First order with respect to Br^- , first order with respect to BrO^-₃ and second order with respect to H⁺ . Hence the overall order of the reaction is equal to 1 + 1 + 2 = 4

2. Order of the reaction with respect to acetaldehyde is \(\frac{3}{2}\) and overall order is also \(\frac{3}{2}\)

Rate constant of reaction, k = 60 s^-1

t_15/16 = ?

Rate constant of first order reaction is given as,

\(k\,=\,\frac{2.303}{t}log\frac{a}{a-x}\) 

or, \(t\,=\,\frac{2.303}{k}log\frac{a}{a-x}\)

When \(\frac{15}{16}th\) of reaction is over,

if a = 1 M, then a - x = \(\frac{1}{16}\) M 

Hence,

\(t_\frac{15}{16}\) = \(\,\frac{2.303}{60\,s^{-1}}log\frac{1}{1-\frac{15}{16}}\)

\(=\,\frac{2.303}{60\,s^{-1}}log\,16\) 

\(=\,\frac{2.303}{60\,s^{-1}}\) x 1.2041s

= 0.046 s

1. Photochemical reaction

2. Order = 0 Molecularity = 2

3. L^-1 mol¹sec^-1

4. Initiation, propagation, termination

a) i) (a + b) – Molecularity of the reaction

(X + y) – Order of the reaction

ii)

b) The order of a reaction can be zero which means that the rate of the reaction is independent of the concentration of the reactants. But, molecularity of a reaction cannot be zero which means that there is no reacting species and hence no reaction is possible.

1. Time in which the concentration of the reactant is reduced to one half of its initial concentration.

2. \(t_{1/2}=\frac{0.693}{k}=\frac{0.693}{5.5\times 10^{-14}s^{-1}}\)\(=\frac{0.693}{5.5}\times10^{14}s\)

T₁ = 293K,

T₂ = 313K 

According to Arrhenius equation,

\(log\frac{k_2}{k_1}=\frac{E_a}{2.303\,R}[\frac{T_2\,-\,T_1}{T_1\,T_2}]\)

∴ E_a = 2.303 R x \([\frac{T_1\,T_2}{T_2\,-\,T_1}]log\frac{k_2}{k_1}\) 

∴ E_a = 2.303 R x 8.314 JK^-1mol^-1 x \([\frac{293\,\times\,313}{313\,-\,293}]log\frac{4}{1}\)

2.303 × 8.314 J K^-1 mol^-1 × 4585.45 K × 0.6021 

= 52863.33 J mol^-1

= 52.86 kJ mol^-1

(1). (a) Order = \(\frac{1}{2}+\frac{3}{2}\) = 2

 (b) Order = \(\frac{3}{2}+(-1)\) = \(\frac{1}{2}\)

(2). For a 1^st order reaction,

k = \(\frac{2.303}{t}log\frac{[R]_0}{[R]}\)

= \(\frac{2.303}{60}log\frac{1.24\,\times\,10^{-2}}{0.20\,\times\,10^{-2}}\)

= \(\frac{2.303}{60}log\,6.2\,min^{-1}\)

k = 0.0304 min^-1

1. First order reaction.

2. 1/4^th life of a first order reaction can be determined as follows.

\(k=\frac{2.303}{t}log\frac{[R]_0}{[R]}\)

When t = t/4, [R] can be taken as \(\frac{[R]_0}{4}\)

\(k=\frac{2.303}{t/4}log\,4\,;\)

\(k=\frac{2.303}{t/4}\times log\,4\)

t_t/4 = \(\frac{2.303\,\times\,log\,4}{k}\)

\(=\frac{2.303\,\times\,0.6020}{k}\) 

\(=\frac{1.386}{k}\)

1. Rate expression :

Rate = \(-\frac{1}{2}\,\frac{d[N_2O_5]}{dt}=\)\(+\frac{1}{4}\,\frac{d[NO_2]}{dt}= + \frac{d[O_2]}{dt}\)

2.

• Increases by 4 times
• Increases by \(\sqrt2\) times. 

Initial rate = k[A][B]²

But [A] = 0.1 M, [B] = 0.2 M and 

k = 2.0 × 10^-6 mol^-2 L² s^-1

∴ Initial rate

= 2.0 × 10^-6 mol^-2 L² s^-1 × 0.1 M × (0.2 M)²

= 8 × 10^-9 M s^-1

From the equation 2A + B → A₂B it is clear that when 2 moles of A are used then 1 mol of B is used in the same time. Therefore, when A has been reduced to 0.06 M , 0.04 M of A and hence 0.02 M of B have reacted.

Thus,

Concentration of A left = [A] = 0.06 M

Concentration of B left = [B] = (0.2 M – 0.02 M) = 0.18 M

Rate = k[A][B]²

= 2.0 × 10^-6 mol^-2 L² s^-1 × 0.06 M × (0.18 M)²

= 3.89 × 10^-9 M s^-1

1. Order is the sum of the powers of the concentration terms in the rate law.

 rate = K [A]^x [B]^y 

∴ Order = x + y

2. According to Arrhenius equation, at a particular temperature for a definite value of activation energy, rate constant,

k ∝ e^-1/T 

3. If average kinetic energy of reactant molecule is equal to threshold energy, then activation energy can be zero.

Initial rate = k[A][B]²

But [A] = 0.1 M, [B] = 0.2 M and k = 2.0 × 10^-6 mol^-2 L² s^-1

Initial rate

2.0 × 10^-6 mol^-2 L² s^-1 × 0.1 M × (0.2 M)²

= 8 × 10^-9 M s^-1

From the equation 2A + B → A₂B it is clear that when 2 moles of A are used then 1 mol of B is used in the same time. Therefore, when A has been reduced to 0.06 M , 0.04 M of A and hence 0.02 M of B have reacted.

Thus,

Concentration of A left = [A] = 0.06 M. Concentration of B left = [B] = (0.2 M – 0.02 M) = 0.18 M

Rate = k[A][B]²

= 2.0 × 10^-6 mol^-2 L² s^-1 × 0.06 M × (0.18 M)²

= 3.89 × 10^-9 M s^-1

(a) Rate of decomposition of N₂O₅.

(b) Rate of formation of O₂.

(c) Unit of rate = mol litre^-1 time^-1.

Transition state or activated complex : 

The configuration of atoms formed from reactant molecules and which is at the peak of barrier in energy profile diagram having maximum potential energy compared to reactants and products is called transition state or activated complex.

The unit of rate constant suggest it to be first order.

Thus rate = K [A]

= 2.1 x 10^-5 x 0.2 

= 4.2 x 10^-6 mol litre^-1 sec^-1

The change in concentration w.r.t. time is called rate of reaction.