Question

if temperature is reduced from 50 degree to 10 degree the rate constant becomes – times of initial value ​

Answer 1

Answer:

For cellular reactions to occur fast enough over short time scales, their activation energies are lowered by molecules called catalysts.

Explanation:

Answer 2

Given: Given that the temperature is reduced from 50 degree to 10 degree.

To find: We have find out the change in rate constant.

Solution:

According to Arrhenius equation

$k = A. {e}^{ \frac{ - E}{R T} }$

Where k is rate constant.

A is a constant and E is energy of activation.

R is Ridberg constant and T is temperature.

The ratio of rate constants for 50 degree temperature and 30 degree temperature is-

$ln(\frac{k_1}{k_2} ) = ( \frac{1}{10} - \frac{1}{50}) \\ ln(\frac{k_1}{k_2} ) = 0.08 \\ (\frac{k_1}{k_2} ) = 1.08$

So, the rate constant becomes 1.08 times of initial.