Question

The rate of a reaction A doubles on increasing the temperature from $300$ to ${\rm{310}}\;{\rm{K}}$. By how much, the temperature of reaction B should be increased from ${\rm{300}}\;{\rm{K}}$ so that rates double if activation energy of reaction B is twice that of reaction A?
A. ${\rm{4}}{\rm{.92}}\;{\rm{K}}$
B. ${\rm{19}}{\rm{.67}}\;{\rm{K}}$
C. ${\rm{2}}{\rm{.45}}\;{\rm{K}}$
D. ${\rm{9}}{\rm{.84}}\;{\rm{K}}$

Answer 1

Hint: We know that the Arrhenius equation is useful for calculating the unknown or final temperature of any reaction. It has temperature, activation energy, rate constant and gas constant. The temperature is usually taken in kelvin.

Complete step by step solution:
We have given data as follows,
The initial temperature of reaction A is ${\rm{300}}\;{\rm{K}}$.
The final temperature of reaction A is ${\rm{310}}\;{\rm{K}}$.
The initial temperature of reaction B is ${\rm{300}}\;{\rm{K}}$.
According to the given statement the rate of the reaction increases which means for reaction A:
${\rm{k'}} = {\rm{2k}}$
As we all know, the Arrhenius equation is needed for the two temperature and rate constants as shown below.
${\mathop{\rm l}\nolimits} {\rm{og}}\dfrac{{{\rm{k'}}}}{{\rm{k}}} = \dfrac{{\rm{E}}}{{{\rm{2}}{\rm{.303R}}}}\left[ {\dfrac{{{\rm{T'}} - {\rm{T}}}}{{{\rm{TT'}}}}} \right]$
Substitute all the respective values in above equation we get,
$
\Rightarrow {\mathop{\rm l}\nolimits} {\rm{og}}\dfrac{{{\rm{2k}}}}{{\rm{k}}} = \dfrac{{\rm{E}}}{{{\rm{2}}{\rm{.303R}}}}\left[ {\dfrac{{310 - 300}}{{310 \times 300}}} \right]\\
\Rightarrow \log 2 = \dfrac{{\rm{E}}}{{{\rm{2}}{\rm{.303R}}}}\left[ {\dfrac{{310 - 300}}{{310 \times 300}}} \right]......\left( 1 \right)
$
Similarly, for equation B the Arrhenius equation is shown below.
${\mathop{\rm l}\nolimits} {\rm{og}}\dfrac{{{\rm{k'}}}}{{\rm{k}}} = \dfrac{{\rm{E}}}{{{\rm{2}}{\rm{.303R}}}}\left[ {\dfrac{{{\rm{T'}} - {\rm{T}}}}{{{\rm{TT'}}}}} \right]$
The given condition in the question for reaction B is shown below.
$
{\rm{k'}} = {\rm{2k}}\\
\Rightarrow {{\rm{E}}_{\rm{a}}}{\rm{(B)}} = {\rm{2}}{{\rm{E}}_{\rm{a}}}{\rm{(A)}}
$
Substitute all the respective values in the above equation.
$
\Rightarrow {\mathop{\rm l}\nolimits} {\rm{og}}\dfrac{{{\rm{2k}}}}{{\rm{k}}} = \dfrac{{{\rm{2E}}}}{{{\rm{2}}{\rm{.303R}}}}\left[ {\dfrac{{{\rm{T'}} - 300}}{{{\rm{300}} \times {\rm{T'}}}}} \right]\\
\Rightarrow \log 2 = \dfrac{{{\rm{2E}}}}{{{\rm{2}}{\rm{.303R}}}}\left[ {\dfrac{{{\rm{T'}} - 300}}{{{\rm{300}} \times {\rm{T'}}}}} \right]......\left( 2 \right)
$
Now, divide the equation (2) by equation (1).
$
\Rightarrow \dfrac{{\log 2}}{{\log 2}} = \dfrac{{\dfrac{{{\rm{2E}}}}{{{\rm{2}}{\rm{.303R}}}}\left[ {\dfrac{{{\rm{T'}} - 300}}{{{\rm{300}} \times {\rm{T'}}}}} \right]}}{{\dfrac{{\rm{E}}}{{{\rm{2}}{\rm{.303R}}}}\left[ {\dfrac{{10}}{{{\rm{300}} \times 310}}} \right]}}\\
\Rightarrow 1 = 2 \times \dfrac{{\left[ {\dfrac{{{\rm{T'}} - 300}}{{{\rm{T'}}}}} \right]}}{{\left[ {\dfrac{1}{{31}}} \right]}}\\
\Rightarrow \dfrac{1}{{62}} = \dfrac{{{\rm{T'}} - 300}}{{{\rm{T'}}}}\\
\Rightarrow {\rm{T'}} - 300 = 0.01613{\rm{T'}}
$
Further solve the above equation we get,
$
\Rightarrow 0.9838{\rm{T'}} = 300\\
\Rightarrow {\rm{T'}} = 304.92\;{\rm{K}}
$
The temperature f reaction which should be increased from ${\rm{300}}\;{\rm{K}}$ by $304.92\;{\rm{K}}$ is $304.92\;{\rm{K}} - {\rm{300}}\;{\rm{K}} = {\rm{4}}{\rm{.92}}\;{\rm{K}}$.
Hence, “the temperature of reaction B should be increased from ${\rm{300}}\;{\rm{K}}$ so that rates doubles if activation energy of the reaction B is twice to that of reaction A” is $4.92\;{\rm{K}}$.

Therefore, the correct option for this given question is A that is $4.92\;{\rm{K}}$.

Note:
The Arrhenius equation usually is affected by the activation energy and temperatures. It generally is used to calculate the missing temperature of the chemical reaction, activation energy and frequency factor.