The activation energies of two reactions are E1 and E2(E1E2) If the temperature of the system is increased from T1 to T2, he rate constant of the reactions changes from k1 to k'1 in the first reaction and k2 to k'2 in the second reaction. Predict which of the following expression is correct?

Question

The activation energies of two reactions are E1 and E2(E1>E2) If the temperature of the system is increased from T1 to T2, he rate constant of the reactions changes from k1 to k'1 in the first reaction and k2 to k'2 in the second reaction. Predict which of the following expression is correct? (A)  (k'1/k1)=(k'2/k2)  (B)  (k'1/k1) >(k'2/k2)  (C)  (k'1/k1) < (k'2/k2)  (D)  (k'1/k1)=(k'2/k2)=1

Tardigrade Admin · Accepted Answer

For first reaction, E1=(2.303RT1T2/(T1-T2)) log (k1/k1) ..(i) For second reaction, E2=(2.303RT1T2/(T1-T2)) log (k2/k2) ..(ii) Given, E1>E2 ⇒ (2.303RT1T2/(T1-T2)) log (k1/k1)>(2.303RT1T2/(T1-T2)) log (k2/k2) ∴ (k1/k1)>(k2/k2)

$\frac{k _{1}^{'}}{k _{1}} = \frac{k _{2}^{'}}{k _{2}}$

$\frac{k _{1}^{'}}{k _{1}} > \frac{k _{2}^{'}}{k _{2}}$

$\frac{k _{1}^{'}}{k _{1}} < \frac{k _{2}^{'}}{k _{2}}$

$\frac{k _{1}^{'}}{k _{1}} = \frac{k _{2}^{'}}{k _{2}} = 1$

$\frac{k _{1}^{'}}{k _{1}} = \frac{k _{2}^{'}}{k _{2}} = 0$

k1​k1′​​=k2​k2′​​

k1​k1′​​>k2​k2′​​

k1​k1′​​<k2​k2′​​

k1​k1′​​=k2​k2′​​=1

k1​k1′​​=k2​k2′​​=0

$\frac{k _{1}^{'}}{k _{1}} = \frac{k _{2}^{'}}{k _{2}}$

$\frac{k _{1}^{'}}{k _{1}} > \frac{k _{2}^{'}}{k _{2}}$

$\frac{k _{1}^{'}}{k _{1}} < \frac{k _{2}^{'}}{k _{2}}$

$\frac{k _{1}^{'}}{k _{1}} = \frac{k _{2}^{'}}{k _{2}} = 1$

$\frac{k _{1}^{'}}{k _{1}} = \frac{k _{2}^{'}}{k _{2}} = 0$