Question

For a reaction, $\frac{K_{t+10}}{K_t}=x$. When temperature is increased from ${10}^0$ C to ${100}^0$ C, rate constant (K) increased by a factor of 512. Then, value of x is:

• A. 1.5
• B. 2.5
• C. 3
• D. 2

Answer 1

The correct option is D 2

Given:

Temperature difference i.e. $\Delta T=T_2-T_1=100-10=90$

$\frac{K_{T_2}}{K_{T_1}}=512$

$\frac{K_t+10}{K_t}=x$

To find $x$?

So, $\frac{K_{T_2}}{K_{T_1}}={\mu }^{\frac{\Delta T}{10}}$

$\mu$ is temperature coefficient of reaction.

On putting all the given values in this we will find:

$512={\mu }^{\frac{\Delta T}{10}}$

$512={\mu }^{\frac{90}{10}}$

$2^9={\mu }^9$

so $\mu =2$

which is $x$ in this question

Option D is correct answer.