Question

For a reaction, $\dfrac{{{K}_{t+10}}}{{{K}_{t}}}$ =x. When temperature is increased from 10 $^{o}C$ to 100 $^{o}C$ , rate constant (K) increases by a factor of 512. Then, value of x is:
A. 1.5
B. 2.5
C. 3
D. 2

Answer 1

There is a relationship between the rate constant and the temperature coefficient of the chemical reaction and it is as follows.
$\dfrac{{{K}_{{{T}_{2}}}}}{{{K}_{T1}}}={{\mu }^{\dfrac{\Delta T}{10}}}$
Here ${{K}_{{{T}_{2}}}}$ = rate constant of the chemical reaction at temperature ${{T}_{2}}$
${{K}_{{{T}_{1}}}}$ = rate constant of the chemical reaction at temperature ${{T}_{1}}$
$\mu$ = temperature coefficient
$\Delta T$ = change in temperature

Complete answer:
- In the question it is asked to calculate the value of the ‘x’ from the given data in the question.
- The change in temperature of the reaction $\Delta T$ = 100-10 = 90.
- Rate constant (K) increased by a factor of 512.
- Means $\dfrac{{{K}_{{{T}_{2}}}}}{{{K}_{T1}}}$ = 512.
- In the question it is also given that $\dfrac{{{K}_{t+10}}}{{{K}_{t}}}$ = x.
- We have to find the value of the x by using the above data with the below formula.
$\dfrac{{{K}_{{{T}_{2}}}}}{{{K}_{T1}}}={{\mu }^{\dfrac{\Delta T}{10}}}$
Here ${{K}_{{{T}_{2}}}}$ = rate constant of the chemical reaction at temperature ${{T}_{2}}$
${{K}_{{{T}_{1}}}}$ = rate constant of the chemical reaction at temperature ${{T}_{1}}$
$\mu$ = temperature coefficient
$\Delta T$ = change in temperature = 90

& \dfrac{{{K}_{{{T}_{2}}}}}{{{K}_{T1}}}={{\mu }^{\dfrac{\Delta T}{10}}} \\\ & 512={{\mu }^{\dfrac{\Delta T}{10}}} \\\ & 512={{\mu }^{\dfrac{90}{10}}} \\\ & {{2}^{9}}={{\mu }^{9}} \\\ & \mu =2 \\\ \end{aligned}$$ \- Here $\mu $ = x in the given data. \- Therefore, the value of x =2. **So, the correct option is D.** **Note:** We should know the change in temperature of the chemical reaction and the value of the rate constant increased at the time of finding the value of the temperature coefficient of the chemical reaction.