Question

On increasing temperature from $\text{200K}$ to $\text{220K}$ rate of reaction A increases by 3 times and the rate of reaction B increases by 9 times then correct the relationship between the activation energy of A and B is:
A) ${{\text{E}}_{\text{A}}}\text{ = 3}{{\text{E}}_{\text{B}}}$
B) $\text{3}{{\text{E}}_{\text{A}}}\text{ = }{{\text{E}}_{\text{B}}}$
C) ${{\text{E}}_{\text{B}}}\text{ = 2}{{\text{E}}_{\text{A}}}$
D) ${{\text{E}}_{\text{A}}}\text{ = 2}{{\text{E}}_{\text{B}}}$

Answer 1

The Arrhenius equation relates the rate constant with the activation energy, absolute temperature, and the pre-exponential factor A. The rate of reaction is directly proportional to the rate constant. The relation between the rates at temperature $\text{200K}$ and $\text{220K}$can be used to establish the relationship between activation energy for reaction A and B.

Complete answer:
The Arrhenius equation is used for the calculations of activation energy.
$\text{ K = A}{{\text{e}}^{\left( \text{-}\dfrac{{{\text{E}}_{\text{a}}}}{\text{RT}} \right)}}$ (1)
Where
${{\text{E}}_{\text{a}}}$ Is Arrhenius activation energy
A is the pre-exponential factor
R is gas constant and T is the absolute temperature in kelvin
According to the question, We know that,
For reaction A
$\text{ }{{\text{T}}_{\text{1}}}\text{=200 K and T2 = 220K}$
Let $\text{ }{{\text{K}}_{\text{1}}}$ be the rate for A and $\text{ }{{\text{K}}_{2}}$be the rate of reaction B
The relation between the rates of reaction A and B is given as,
$\text{3 (}{{\text{r}}_{\text{1}}}{{\text{)}}_{^{^{_{\text{A}}}}}}\text{= }{{\left( {{\text{r}}_{\text{2}}} \right)}_{\text{A}}}$

& \text{where,} \\\ & {{\text{(}{{\text{r}}_{\text{1}}}\text{)}}_{^{^{_{\text{A}}}}}}\text{= rate of reaction A at }{{\text{T}}_{\text{1}}}\text{ K} \\\ & {{\left( {{\text{r}}_{\text{2}}} \right)}_{\text{A}}}\text{= rate of reaction A at }{{\text{T}}_{\text{2}}}\text{ K } \\\ \end{aligned}$$ Since we know that, $\begin{aligned} & \text{Rate of reaction }\propto \text{ Rate constant } \\\ & \Rightarrow \text{ 3 (}{{\text{K}}_{\text{1}}}{{\text{)}}_{\text{A}}}\text{=(}{{\text{K}}_{\text{2}}}{{\text{)}}_{\text{A}}} \\\ \end{aligned}$ Where, ${{\text{K}}_{\text{1}}}$and ${{\text{K}}_{\text{2}}}$are the rate constant at temperature ${{\text{T}}_{\text{1}}}$ and $${{\text{T}}_{\text{2}}}$$ On further simplification we get, $\text{ }\dfrac{{{\text{(}{{\text{K}}_{\text{1}}}\text{)}}_{\text{A}}}}{{{\text{(}{{\text{K}}_{\text{2}}}\text{)}}_{\text{A}}}}\text{ = }\dfrac{\text{1}}{\text{3}}\text{ }$ (2) Now using (1) and (2) we get, $$\text{ }\dfrac{{{\text{(}{{\text{K}}_{\text{1}}}\text{)}}_{\text{A}}}}{{{\text{(}{{\text{K}}_{\text{2}}}\text{)}}_{\text{A}}}}\text{=}\dfrac{1}{3}\text{ = }\dfrac{\text{A}{{\text{e}}^{\left( \text{-}\dfrac{{{\text{E}}_{\text{a}}}}{\text{R}{{\text{T}}_{\text{1}}}} \right)}}}{\text{A}{{\text{e}}^{\left( \text{-}\dfrac{{{\text{E}}_{\text{a}}}}{\text{R}{{\text{T}}_{\text{2}}}} \right)}}}\text{ }$$ $$\begin{aligned} & \Rightarrow \dfrac{1}{3}\text{= }\dfrac{\text{A}{{\text{e}}^{\left( \text{-}\dfrac{{{\text{E}}_{\text{a}}}}{\text{R}{{\text{T}}_{\text{1}}}} \right)}}}{\text{A}{{\text{e}}^{\left( \text{-}\dfrac{{{\text{E}}_{\text{a}}}}{\text{R}{{\text{T}}_{\text{2}}}} \right)}}} \\\ & \Rightarrow \dfrac{1}{3}\text{= }{{\text{e}}^{\left( \text{-}\dfrac{{{\text{E}}_{\text{a}}}}{\text{R}{{\text{T}}_{\text{1}}}} \right)}}\text{ }{{\text{e}}^{\left( \dfrac{{{\text{E}}_{\text{a}}}}{\text{R}{{\text{T}}_{\text{2}}}} \right)}} \\\ & \Rightarrow \dfrac{1}{3}\text{= }{{\text{e}}^{\text{-}\dfrac{{{\text{E}}_{\text{a}}}}{\text{R}}\left( \dfrac{1}{{{\text{T}}_{\text{2}}}}-\dfrac{1}{{{\text{T}}_{\text{1}}}} \right)}} \\\ \end{aligned}$$ Let us substitute the value for temperature. we get $$\begin{aligned} & \Rightarrow \dfrac{1}{3}\text{= }{{\text{e}}^{\text{-}\dfrac{{{\text{E}}_{\text{a}}}}{\text{R}}\left( \dfrac{1}{220}-\dfrac{1}{200} \right)}} \\\ & \Rightarrow \dfrac{1}{3}\text{= }{{\text{e}}^{\text{ }\dfrac{{{\text{E}}_{\text{a}}}}{\text{R}}\text{ }\left( \dfrac{20}{220\text{ }\times \text{ }200} \right)}} \\\ & \Rightarrow \dfrac{1}{3}\text{= }{{\text{e}}^{\text{ }\dfrac{{{\text{E}}_{\text{a}}}}{\text{R}}\text{ }\left( \dfrac{20}{220\text{ }\times \text{ }200} \right)}} \\\ \end{aligned}$$ Take a logarithm on both sides of the equation. $$\begin{aligned} & \Rightarrow \log \left( \dfrac{1}{3} \right)\text{= }\dfrac{{{\text{E}}_{\text{a}}}}{\text{R}}\text{ }\left( \dfrac{20}{220\text{ }\times \text{ }200} \right) \\\ & \Rightarrow \log \left( \dfrac{1}{3} \right)\text{= }\dfrac{{{\text{E}}_{\text{a}}}}{\text{R}}\text{ }\left( \dfrac{1}{2200} \right) \\\ & \text{ Rearrange equation to get the value of }{{\text{E}}_{\text{a}}}, \\\ & {{\text{E}}_{\text{a}}}=2200\text{ R }\log \left( \dfrac{1}{3} \right) \\\ \end{aligned}$$ For reaction B $\text{ }{{\text{T}}_{\text{1}}}\text{=200 K and }{{\text{T}}_{\text{2}}}\text{ = 220K}$ Let $\text{ }{{\text{K}}_{\text{1}}}$ be the rate for A and $\text{ }{{\text{K}}_{2}}$be the rate of reaction B The relation between the rates of reaction A and B is given as, $$\text{9 (}{{\text{r}}_{\text{1}}}{{\text{)}}_{^{^{\text{B}}}}}\text{= }{{\left( {{\text{r}}_{\text{2}}} \right)}_{\text{B}}}$$ $$\begin{aligned} & \text{where,} \\\ & {{\text{(}{{\text{r}}_{\text{1}}}\text{)}}_{^{^{_{\text{B}}}}}}\text{= rate of reaction B at }{{\text{T}}_{\text{1}}}\text{ K} \\\ & {{\left( {{\text{r}}_{\text{2}}} \right)}_{\text{B}}}\text{= rate of reaction B at }{{\text{T}}_{\text{2}}}\text{ K } \\\ \end{aligned}$$ Since we know that, $$\begin{aligned} & \text{Rate of reaction }\propto \text{ Rate constant } \\\ & \Rightarrow \text{ 9 (}{{\text{K}}_{\text{1}}}{{\text{)}}_{\text{B}}}\text{=(}{{\text{K}}_{\text{2}}}{{\text{)}}_{\text{B}}} \\\ \end{aligned}$$ Where, ${{\text{K}}_{\text{1}}}$and ${{\text{K}}_{\text{2}}}$are the rate constant at temperature ${{\text{T}}_{\text{1}}}$ and $${{\text{T}}_{\text{2}}}$$ On further simplification we get, $$\text{ }\dfrac{{{\text{(}{{\text{K}}_{\text{1}}}\text{)}}_{\text{B}}}}{{{\text{(}{{\text{K}}_{\text{2}}}\text{)}}_{\text{B}}}}\text{ = }\dfrac{\text{1}}{9}\text{ }$$ (2) By following the similar procedure as above we get, $$\text{ }\dfrac{{{\text{(}{{\text{K}}_{\text{1}}}\text{)}}_{\text{B}}}}{{{\text{(}{{\text{K}}_{\text{2}}}\text{)}}_{\text{B}}}}\text{=}\dfrac{\text{1}}{\text{9}}\text{ = }\dfrac{\text{A}{{\text{e}}^{\left( \text{-}\dfrac{{{\text{E}}_{\text{b}}}}{\text{R}{{\text{T}}_{\text{1}}}} \right)}}}{\text{A}{{\text{e}}^{\left( \text{-}\dfrac{{{\text{E}}_{\text{b}}}}{\text{R}{{\text{T}}_{\text{2}}}} \right)}}}\text{ }$$ $$\Rightarrow \dfrac{1}{9}\text{= }{{\text{e}}^{\text{-}\dfrac{{{\text{E}}_{\text{b}}}}{\text{R}}\left( \dfrac{1}{{{\text{T}}_{\text{2}}}}-\dfrac{1}{{{\text{T}}_{\text{1}}}} \right)}}$$ Take a logarithm on both sides of the equation. $$\begin{aligned} & \Rightarrow \log \left( \dfrac{1}{9} \right)\text{= }\dfrac{{{\text{E}}_{\text{b}}}}{\text{R}}\text{ }\left( \dfrac{20}{220\text{ }\times \text{ }200} \right) \\\ & \Rightarrow \log \left( \dfrac{1}{9} \right)\text{= }\dfrac{{{\text{E}}_{\text{b}}}}{\text{R}}\text{ }\left( \dfrac{1}{2200} \right) \\\ & \text{ Rearrange equation to get the value of }{{\text{E}}_{\text{b}}}, \\\ & {{\text{E}}_{\text{b}}}=2200\text{ R }\log \left( \dfrac{1}{9} \right)\text{ (B)} \\\ \end{aligned}$$ Now from the A and B, we get $\begin{aligned} & \text{ }\dfrac{{{\text{E}}_{\text{a}}}}{{{\text{E}}_{\text{b}}}}\text{ = }\dfrac{\text{log}\left( \dfrac{\text{1}}{\text{3}} \right)}{\text{log}\left( \dfrac{\text{1}}{\text{9}} \right)} \\\ & \Rightarrow \text{ }\dfrac{{{\text{E}}_{\text{a}}}}{{{\text{E}}_{\text{b}}}}\text{ = }\dfrac{\text{1}}{\text{2}} \\\ & \Rightarrow \text{2}{{\text{E}}_{\text{a}}}\text{ = }{{\text{E}}_{\text{b}}} \\\ \end{aligned}$ Thus the relation between activation energy of A and B is $\text{2}{{\text{E}}_{\text{A}}}\text{ = }{{\text{E}}_{\text{B}}}$ **Thus, (C) is the correct option.** **Note:** The Arrhenius equation relates the rate constant with the temperature and activation energy. However rate of reaction is directly proportional to the rate constant. Thus whenever such a problem is asked, always make relation and then divide them by each other .This cancels out the extra terms and makes the equation simple and easy to solve.