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Question: When \[{\text{Ln k}}\] is plotted against \[1/{\text{T}}\], the slope was found to be \[ - 10.7 \tim...

When Ln k{\text{Ln k}} is plotted against 1/T1/{\text{T}}, the slope was found to be −10.7×103 k - 10.7 \times {10^3}{\text{ k}}, activation energy for the reaction would be:
A.−78.9 KJ mol−1 - 78.9{\text{ KJ mo}}{{\text{l}}^{ - 1}}
B.2.26 KJ mol−12.26{\text{ KJ mo}}{{\text{l}}^{ - 1}}
C.88.9 KJ mol−188.9{\text{ KJ mo}}{{\text{l}}^{ - 1}}
D.10.7 KJ mol−1{\text{10}}{\text{.7 KJ mo}}{{\text{l}}^{ - 1}}

Explanation

Solution

To know the relation between activation energy, temperature and k, we need to use the Arrhenius equation. Taking antilog both sides we will have the required relation. On comparing the equation with the equation for straight lines we will get the value of activation energy.
Formula used:
k=Ae−Ea/RT{\text{k}} = {\text{A}}{{\text{e}}^{ - {{\text{E}}_{\text{a}}}{\text{/RT}}}} here k is rate constant. Ea{{\text{E}}_{\text{a}}} is activation energy, R is universal gas constant, T is temperature and e is exponential.

Complete step by step answer:
Arrhenius gave the equation in his activation theory which is as follow:
k=Ae−Ea/RT{\text{k}} = {\text{A}}{{\text{e}}^{ - {{\text{E}}_{\text{a}}}{\text{/RT}}}}
We will multiply both the sides with natural log.
ln k=ln (Ae−Ea/RT){\text{ln k}} = {\text{ln (A}}{{\text{e}}^{ - {{\text{E}}_{\text{a}}}{\text{/RT}}}})
ln k=ln A+ln e−Ea/RT{\text{ln k}} = {\text{ln A}} + {\text{ln }}{{\text{e}}^{ - {{\text{E}}_{\text{a}}}{\text{/RT}}}}
Exponential and natural log are reciprocal to each other and hence they will cancel out each other.
ln k=ln A−EaR×1T{\text{ln k}} = {\text{ln A}} - \dfrac{{{{\text{E}}_{\text{a}}}}}{{\text{R}}} \times \dfrac{1}{{\text{T}}}
The above equation represents the linear straight line equation, y=mx+c{\text{y}} = {\text{mx}} + {\text{c}}. When a graph is plotted against y and x then m is the slope and c is the intercept.
In our question y is k and x is 1T\dfrac{1}{{\text{T}}} so the slope becomes −EaR\dfrac{{ - {{\text{E}}_{\text{a}}}}}{{\text{R}}}. The value of slope is given to us that is −10.7×103 k - 10.7 \times {10^3}{\text{ k}}.
We will compare both values. We will get,
−EaR=−10.7×103 k\dfrac{{ - {{\text{E}}_{\text{a}}}}}{{\text{R}}} = - 10.7 \times {10^3}{\text{ k}}
⇒Ea=10.7×103 K × R\Rightarrow {{\text{E}}_{\text{a}}} = 10.7 \times {10^3}{\text{ K }} \times {\text{ R}}
The value of R is constant that is 8.314 ×10−3 kJ K−1 mol−18.314{\text{ }} \times {\text{1}}{{\text{0}}^{ - 3}}{\text{ kJ }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}.
Ea=10.7×103 K × 8.314 ×10−3kJ K−1 mol−1{{\text{E}}_{\text{a}}} = 10.7 \times {10^3}{\text{ K }} \times {\text{ }}8.314{\text{ }} \times {10^{ - 3}}{\text{kJ }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}
⇒Ea=88.9kJ mol−1\Rightarrow {{\text{E}}_{\text{a}}} = 88.9{\text{kJ mo}}{{\text{l}}^{ - 1}}

Hence, the correct option is C.

Note:
According to Arrhenius theory the activation energy is independent of temperature. In actual practice activation energy also varies with temperature. The higher the temperature, higher is the impact of temperature on activation energy. Activation energy is the amount of energy required by the reactant molecules to change into a product. k is the rate constant which is a constant at particular temperature. When temperature changes the value of the rate constant also changes.