Question

$A\to B;K_A={10}^{15}{e}^{\frac{-2000}{T}}$

$C\to D;K_C={10}^{14}{e}^{\frac{-1000}{T}}$
Temperature T in terms of K at which $K_A=K_C$ is :

• A. $1000K$
• B. $2000K$
• C. $\frac{2000}{2.303}K$
• D. $\frac{1000}{2.303}K$

Answer 1

The correct option is D $\frac{1000}{2.303}K$

The rate constant of two reactions will be equal, $K_A=K_C$.

Hence,

${10}^{15}{e}^{\frac{-2000}{T}}={10}^{14}{e}^{\frac{-1000}{T}}$

$ln{10}^{15}+\frac{-2000}{T}=ln{10}^{14}+\frac{-1000}{T}$

$\therefore 15\times \left(2.303\right)-14\times \left(2.303\right)=\frac{2000}{T}-\frac{1000}{T}$

Thus, the temperature at which the rate constants of two reactions will be equal is $T=\frac{1000}{2.303}K.$

Hence, the correct option is $\text{D}$