Question

Question:
The rate of reaction becomes two times for every 10-degree rise in temperature. How much will the rate of reaction will increase, when the temperature is increased from 30-degree C to 80-degree C?
A.16
B.32
C.64
D.128

Answer 1

Hint: The rate of reaction increases with the rise in temperature. By using $\dfrac{{{r_2}}}{{{r_1}}} = {2^n}$ , we will find the value of ‘n’.

Complete step by step answer:
Given,
The rate of reaction becomes twice for every $10^\circ $ rise in temperature,
The temperature is increased from $30^\circ C$ to $80^\circ C$ , thus the temperature is raised by $50^\circ C$
Thus, the temperature coefficient of the reaction = 2 when the temperature is increased by $50^\circ C$
As we know, $\dfrac{{{r_2}}}{{{r_1}}} = {2^n}$
Where, ${r_2}$ = Rate at ${T_2}$ $(80^\circ C)$
 ${r_1}$ =Rate at ${T_1}$ $(30^\circ C)$
 $n = \dfrac{{\Delta T}}{{T1}}$
when temperature increases from $30^\circ C$ to $80^\circ C$ , change in temperature
  $n = \dfrac{{80 - 30}}{{10}}$ = 5
So, number of times the rate of reaction doubles is 5
Thus, the rate of reaction increases by ${2^5}$ = 32 times
Therefore, the correct answer is option (B).

Note: Rate is dependent on rate constant as the rate law expression states, rate is directly proportional to rate constant and temperature.
According to Arrhenius equation,
K = A * ${e^{(\dfrac{{ - energy}}{{RT}})}}$
Where, K = rate constant
A = Arrhenius constant
T = temperature
If we increase the temperature the rate of reaction increases because the particles gain more kinetic energy i.e. speed of the particles is proportional to the square root of its kinetic energy and therefore more effective collisions can be made (increase the number of collisions). Heat increases the average energy of the molecules. But many reactions are not simple and their rate may actually decrease with increasing temperature.