Question

Velocity constant of a reaction at $290K$ was found to be ${3.2\times 10}^{-3}$. At $310K$ it will be about?

• A. ${1.28\times 10}^{-2}$
• B. ${9.6\times 10}^{-3}$
• C. ${6.4\times 10}^{-3}$
• D. ${3.2\times 10}^{-4}$

Answer 1

The correct option is C ${6.4\times 10}^{-3}$

From the given question there is a increase in temperature is ${20}^{0}C$.

As the temperature increases ${20}^{0}C$, the velocity constant doubles.

So, the velocity constant at $310K$ is $6.4\times {10}^{-3}$.