Question

For a chemical reaction the specific rate constant at $\text{600K}$ is $\text{1}\text{.60 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{8}}}{{\text{s}}^{\text{-1}}}$, calculate the rate constant for the reaction at $\text{700K}$, if energy of activation for the reaction is $\text{2}\text{.209KJmo}{{\text{l}}^{\text{-1}}}$.

Answer 1

On increasing temperature rate of reaction increases whether the reaction is exothermic or endothermic when temperature increases $\text{KE}$ of molecules increases, number of activated molecules increases. Thus the rate constant is increased. Arrhenius proposed a mathematical expression to give a quantitative relation between the rate constant and temperature.
$\begin{aligned} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{k =}\,\,\text{A}\text{.}{{\text{e}}^{\text{-}\dfrac{{{\text{E}}_{\text{a}}}}{\text{RT}}}}..........(i) \\\ & \text{where,}\,\,\text{A}\,-\,\text{frequency}\,\text{factor or Arrhenius constant} \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\text{E}}_{\text{a}}}-\,\text{activation}\,\text{energy} \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{R -}\,\,\text{gas}\,\text{constant} \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{T}\,\,-\,\text{temperature in K} \\\ \end{aligned}$
- Rate constant is a proportionality constant which is represented by ‘k’, the value of ‘k’ depends on the particular reaction which is being studied and temperature at which reaction takes place.
- Activation energy of a reaction is represented by ${{\text{E}}_{\text{a}}}$. It is the minimum extra amount of energy required by reactant molecules for converting into product, or the additional energy which is required to attain threshold energy is called activation energy.

Complete Solution :
We will calculate the rate constant with the help of Arrhenius equation –
Suppose at temperature ${{\text{T}}_{\text{1}}}\text{K}$ and ${{\text{T}}_{2}}\text{K}$ the rate constants of the reaction are ${{\text{K}}_{\text{1}}}$ and ${{\text{K}}_{2}}$ respectively so after applying equation (i) we get ${{\text{k}}_{\text{1}}}\text{=}\,\text{A}{{\text{e}}^{\text{-}\dfrac{{{\text{E}}_{\text{a}}}}{\text{R}{{\text{T}}_{\text{1}}}}}}........\text{(ii)}$
${{\text{k}}_{\text{2}}}\text{=}\,\text{A}{{\text{e}}^{\text{-}\dfrac{{{\text{E}}_{\text{a}}}}{\text{R}{{\text{T}}_{\text{2}}}}}}........\text{(iii)}$
After dividing equation (iii) by equation (ii) we get

& \dfrac{{{\text{k}}_{\text{2}}}}{{{\text{k}}_{\text{1}}}}\text{=}\,\frac{{{\text{e}}^{\text{-}\dfrac{{{\text{E}}_{\text{a}}}}{\text{R}{{\text{T}}_{\text{2}}}}}}}{{{\text{e}}^{\text{-}\dfrac{{{\text{E}}_{\text{a}}}}{\text{R}{{\text{T}}_{\text{1}}}}}}} \\\ & \text{after}\,\text{taking}\,\text{log}\,\text{on}\,\text{both}\,\text{side} \\\ & \text{log(}\dfrac{{{\text{k}}_{\text{2}}}}{{{\text{k}}_{\text{1}}}}\text{) = }\,\text{log}\,\text{(}{{\text{e}}^{\text{-}\dfrac{{{\text{E}}_{\text{a}}}}{\text{R}}\text{(}\dfrac{\text{1}}{{{\text{T}}_{\text{2}}}}\text{-}\dfrac{\text{1}}{{{\text{T}}_{\text{1}}}}\text{)}}}\text{)} \\\ & \text{log(}\dfrac{{{\text{k}}_{\text{2}}}}{{{\text{k}}_{\text{1}}}}\text{) =}\,\,\dfrac{{{\text{E}}_{\text{a}}}}{\text{R}}\text{(}\dfrac{\text{1}}{{{\text{T}}_{\text{1}}}}\text{-}\dfrac{\text{1}}{{{\text{T}}_{\text{2}}}}\text{)}.......\text{(iv)} \\\ & \\\ \end{aligned}$$ After putting ${{\text{T}}_{\text{1}}}\text{= 600K,}\,\,{{\text{K}}_{1}}=\,1.60\times {{10}^{-6}}{{\text{s}}^{\text{-1}}}\,\,{{\text{T}}_{2}}\text{= 700K,}\,{{\text{E}}_{\text{a}}}\text{= 2}\text{.209 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{3}}}{\scriptstyle{}^{\text{J}}/{}_{\text{mol}}}\,\,\text{and}\,\text{R = 8}\text{.314J}{{\text{K}}^{\text{-1}}}\text{mo}{{\text{l}}^{\text{-1}}}$ In equation (IV) we will calculate the value of ${{\text{k}}_{\text{2}}}$ $$\begin{aligned} & \text{log(}\dfrac{{{\text{k}}_{\text{2}}}}{{{\text{k}}_{\text{1}}}}\text{) =}\,\,\dfrac{{{\text{E}}_{\text{a}}}}{\text{R}}\text{(}\dfrac{\text{1}}{{{\text{T}}_{\text{1}}}}\text{-}\dfrac{\text{1}}{{{\text{T}}_{\text{2}}}}\text{)} \\\ & \text{log(}\dfrac{{{\text{k}}_{\text{2}}}}{{{\text{k}}_{\text{1}}}}\text{) =}\,\dfrac{\text{2209}}{\text{8}\text{.314}}\text{(}\dfrac{\text{1}}{\text{600}}\text{-}\frac{\text{1}}{\text{700}}\text{)} \\\ & \text{log(}\dfrac{{{\text{k}}_{\text{2}}}}{{{\text{k}}_{\text{1}}}}\text{)=}\,\dfrac{\text{2209 }\\!\\!\times\\!\\!\text{ 100}}{\text{8}\text{.314 }\\!\\!\times\\!\\!\text{ 6 }\\!\\!\times\\!\\!\text{ 7}} \\\ & \text{log(}\dfrac{{{\text{k}}_{\text{2}}}}{{{\text{k}}_{\text{1}}}}\text{)=}\,\text{0}\text{.63} \\\ & \text{after}\,\text{taking}\,\text{antilog}\,\text{in}\,\text{both}\,\text{sides-} \\\ & \dfrac{{{\text{k}}_{\text{2}}}}{\text{1}\text{.6 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{6}}}}\text{=}\,\,\text{antilog (0}\text{.63)} \\\ & {{\text{k}}_{\text{2}}}\,\,\text{=}\,\,\text{1}\text{.87 }\\!\\!\times\\!\\!\text{ 1}\text{.6 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{6}}} \\\ & {{\text{k}}_{\text{2}}}\,\,\text{=}\,\text{3 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{6}}}{{\text{s}}^{\text{-1}}} \\\ \end{aligned}$$ **Note:** -Unit of rate constant also depends on the order of reaction so, ${{\text{s}}^{\text{-1}}}$ represents the first order of reaction. Temperature must be present in kelvin. -${{\text{e}}^{\text{-}\dfrac{{{\text{E}}_{\text{a}}}}{\text{RT}}}}$ is known as Boltzman factor, it is the fraction of rate constant and Arrhenius constant. -Rate constant value increases with increasing temperature.