Question

For a reaction, $\dfrac{{{{\text{K}}_{{\text{(t + 10)}}}}}}{{{{\text{K}}_{{\text{(t)}}}}}}{\text{ = x}}$ . When temperature is increased from ${\text{10}}^\circ {\text{C to 100}}^\circ {\text{C}}$ , rate constant (${\text{K}}$) increased by a factor of $512$ . Then, value of ${\text{x}}$ is
A.${\text{1}}{\text{.5}}$
B.${\text{2}}{\text{.5}}$
C.${\text{3}}$
D.${\text{2}}$

Answer 1

We have to remember that the rate of the reaction is an important factor for the study of reaction. The rate of reaction is an important concept for the chemical kinetics. Rate of the reaction depends on the concentration of the reactant. The rate of reaction is also calculated by using the concentration of the product in the chemical reaction. Depending on the concentration, the sign of the rate will change.
Formula used:
For a reaction
$\dfrac{{{{\text{K}}_{{\text{(t + 10)}}}}}}{{{{\text{K}}_{{\text{(t)}}}}}}{\text{ = x}}$
Here, ${\text{K}}$ is proportionality constant, known as rate constant.
t is temperature coefficient of this reaction.
x is temperature coefficient of this reaction.
Temperature difference,
$\Delta {\text{T = }}{{\text{T}}_{\text{2}}}{\text{ - }}{{\text{T}}_{\text{1}}}$
Here, Temperature difference is $\Delta {\text{T}}$ .
The initial temperature is ${{\text{T}}_{\text{1}}}$ .
The final temperature is ${{\text{T}}_{\text{2}}}$ .
$\dfrac{{{{\text{K}}_{{{\text{T}}_{\text{2}}}}}}}{{{{\text{K}}_{{{\text{T}}_{\text{1}}}}}}}{\text{ = }}{{\mu }^{\dfrac{{\Delta {\text{T}}}}{{{\text{10}}}}}}$
Here,Temperature difference is $\Delta {\text{T}}$ .
The rate constant of the initial temperature ${{\text{T}}_{\text{1}}}$ is ${{\text{K}}_{{{\text{T}}_{\text{1}}}}}$
The rate constant of the final temperature ${{\text{T}}_2}$ is ${{\text{K}}_{{{\text{T}}_2}}}$
The temperature coefficient of reaction is $\mu$ .

Complete step by step answer:
The given data is,
$\dfrac{{{{\text{K}}_{{\text{(t + 10)}}}}}}{{{{\text{K}}_{{\text{(t)}}}}}}{\text{ = x}}$
Temperature is increased from ${\text{10}}^\circ {\text{C to 100}}^\circ {\text{C}}$
The rate constant (${\text{K}}$) increased by a factor is $512$ .
The initial temperature ${{\text{T}}_{\text{1}}}$ is ${\text{10}}^\circ {\text{C}}$
The final temperature ${{\text{T}}_{\text{2}}}$ is ${\text{100}}^\circ {\text{C}}$
Temperature difference,
$\Delta {\text{T = }}{{\text{T}}_{\text{2}}}{\text{ - }}{{\text{T}}_{\text{1}}}$
${\text{ = 100}}^\circ {\text{C - 10}}^\circ {\text{C = 90}}^\circ {\text{C}}$
Temperature difference $\Delta {\text{T}}$ is ${\text{90}}^\circ {\text{C}}$ .
$\dfrac{{{{\text{K}}_{{{\text{T}}_{\text{2}}}}}}}{{{{\text{K}}_{{{\text{T}}_{\text{1}}}}}}}{\text{ = }}512$
$\dfrac{{{{\text{K}}_{{{\text{T}}_{\text{2}}}}}}}{{{{\text{K}}_{{{\text{T}}_{\text{1}}}}}}}{\text{ = }}{{{\mu }}^{\dfrac{{\Delta {\text{T}}}}{{{\text{10}}}}}}$
We substitute the known values in above formula,
$512{\text{ = }}{{\mu }^{\dfrac{{90}}{{{\text{10}}}}}}$
$\Rightarrow 512{\text{ = }}{{{\mu }}^{\dfrac{9}{{\text{1}}}}}$
$\Rightarrow {2^9}{\text{ = }}{{{\mu }}^9}$
$\mu = 2$
The temperature coefficient of reaction is $\mu$ .
For this reaction x is temperature coefficient.
$\mu = x$
$x = 2$
According to the above calculation, For a reaction, $\dfrac{{{{\text{K}}_{{\text{(t + 10)}}}}}}{{{{\text{K}}_{{\text{(t)}}}}}}{\text{ = x}}$ . When temperature is increased from ${\text{10}}^\circ {\text{C to 100}}^\circ {\text{C}}$ , rate constant (${\text{K}}$) increased by a factor of $512$ . Then, value of ${\text{x}}$is ${\text{2}}$ .

Hence, option ${\text{D}}$ is correct.

Note:
We have to know that the rate of the reaction depends on the initial concentration of the reactants in the chemical reaction. The rate of the reaction is used to measure decrease in the concentration of reactant and increase the concentration of the product in the reaction. The order of reaction may be fractional or integer or zero. There are five factors that affect the rate of the reaction. Temperature is one of the five factors that affect the rate of the reaction.