Question

For $X\to Y,\frac{k_{t+10}}{k_t}=3$, If the rate constant at $300k$ is $Q{min}^{-1}$, at what temperature rate constant becomes $9Q$ ${min}^{-1}$?

• A. ${47}^{o}C$
• B. ${320}^{o}C$
• C. $280K$
• D. $\sqrt{9\times 300}K$

Answer 1

The correct option is A ${47}^{o}C$

Let $t=300K$

Then rate constant $k_t=Q{min}^{-1}$

Given $⟶\frac{k_{t+10}}{k_t}=3\frac{k_{300+10}}{k_t}=\frac{k_{310}}{k_{300}}=3k_{310}={3.k}_{300}⟶\left(1\right)$

Similarly $k_{320}={3k}_{310}⟶\left(2\right)$

Substituting value of $k_{310}$ in $\left(2\right),$

We get

$k_{320}=3.\left(3k_{300}\right)=9Q.$

Therefore value of rate constant becomes

9Q at $320K={47}^{o}C.$

Therefore answer is [A]