Question

$\begin{align} & \text{ }\left[ \text{A} \right] \\ & \text{ (mol }{{\text{L}}^{-1}}) \\ \end{align}$	$\begin{align} & \text{ }\left[ \text{B} \right] \\ & \text{ (mol }{{\text{L}}^{-1}}) \\ \end{align}$	Initial rate $\text{ (mol }{{\text{L}}^{-1}}\text{ }{{\text{s}}^{-1}})$ at
		$\text{ 300K }$	$\text{ 320K }$
$\text{ 2}\text{.5 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}}\text{ }$	$\text{ 3}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}}\text{ }$	$\text{ 5}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}}\text{ }$	$\text{ 2}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{ }$
$\text{ 5}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}}\text{ }$	$\text{ 6}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}}\text{ }$	$\text{ 4}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{ }$	$\text{ }-\text{ }$
$\text{ 1}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{ }$	$\text{ 6}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}}\text{ }$	$\text{ 1}\text{.6 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\text{ }$	$\text{ }-\text{ }$

a) Calculate the order of reaction with respect to A and with respect to B.
b) Calculate the rate constant at$\text{ 300K }$.

Answer 1

Hint: The order of a reaction with respect to the participating reactants can be calculated by various methods. One of the methods is differential rate expression. According to which, if \[\text{ }{{\left( \text{Rate} \right)}_{1}}\text{ }\]and \[{{\left( \text{Rate} \right)}_{2}}\] are the rate constant and $\text{ }{{\text{C}}_{\text{1}}}\text{ }$ and $\text{ }{{\text{C}}_{2}}\text{ }$of reactant A .Then order of reaction with respect to A is,
$\text{ ln}\dfrac{{{\left( \text{Rate} \right)}_{1}}}{{{\left( \text{Rate} \right)}_{2}}}\text{ = n ln}\dfrac{{{\text{C}}_{\text{1}}}}{{{\text{C}}_{2}}}\text{ }$

Complete answer:
We know that the, rate of reaction is given as follows,
$\text{ Rate = K }{{\left[ \text{A} \right]}^{\text{m}}}{{\left[ \text{B} \right]}^{\text{n}}}\text{ }$
Thus, rate of reaction for set 1, set 2 and set 3 are written as follows,
$\text{ }{{\left( \text{Rate} \right)}_{\text{1}}}\text{ = K }{{\left[ {{\text{A}}_{\text{1}}} \right]}^{\text{m}}}{{\left[ {{\text{B}}_{\text{1}}} \right]}^{\text{n}}}\text{ }$ , $\text{ }{{\left( \text{Rate} \right)}_{2}}\text{ = K }{{\left[ {{\text{A}}_{2}} \right]}^{\text{m}}}{{\left[ {{\text{B}}_{2}} \right]}^{\text{n}}}\text{ }$ and$\text{ }{{\left( \text{Rate} \right)}_{3}}\text{ = K }{{\left[ {{\text{A}}_{3}} \right]}^{\text{m}}}{{\left[ {{\text{B}}_{3}} \right]}^{\text{n}}}\text{ }$ .The order of the reaction with respect to A and B at the concertation can be calculated by taking the ratio of the rate constant of the reaction for various set.
Part A) Lets first consider the rate constant for set 2 and set 3.Divide the rate constant for the set 1 by the set 3.We have ,
$\text{ }\dfrac{{{\text{(Rate)}}_{2}}}{{{\text{(Rate)}}_{3}}}\text{ = }\dfrac{\text{K }{{\left[ {{\text{A}}_{2}} \right]}^{\text{m}}}{{\left[ {{\text{B}}_{2}} \right]}^{\text{n}}}\text{ }}{\text{K }{{\left[ {{\text{A}}_{3}} \right]}^{\text{m}}}{{\left[ {{\text{B}}_{3}} \right]}^{\text{n}}}\text{ }}$
Let’s substitute the values from the table. We have,
$\text{ }\dfrac{{{\text{(Rate)}}_{2}}}{{{\text{(Rate)}}_{3}}}\text{ = }\dfrac{\text{ }{{\left[ {{\text{A}}_{\text{2}}} \right]}^{\text{m}}}{{\left[ {{\text{B}}_{\text{2}}} \right]}^{\text{n}}}\text{ }}{\text{ }{{\left[ {{\text{A}}_{\text{3}}} \right]}^{\text{m}}}{{\left[ {{\text{B}}_{\text{3}}} \right]}^{\text{n}}}\text{ }}\text{= }\dfrac{{{\left( \text{5}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}} \right)}^{\text{m}}}{{\left( \text{6}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}} \right)}^{\text{n}}}}{{{\left( \text{1}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}} \right)}^{\text{m}}}{{\left( \text{6}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}} \right)}^{\text{n}}}}\text{ = }{{\left( \dfrac{\text{5}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}}}{\text{1}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}} \right)}^{\text{m}}}=\text{ }{{\left( \dfrac{1}{2} \right)}^{\text{m}}}\text{ }$ (1)
The ratio of rate constant for set 2 and set 3 is equal to,
$\text{ }\dfrac{{{\text{(Rate)}}_{2}}}{{{\text{(Rate)}}_{3}}}\text{ = }\dfrac{\left( 4.0\times {{10}^{-3}} \right)}{\left( 1.6\times {{10}^{-2}} \right)}\text{ = }\left( \dfrac{1}{4} \right)\text{ }$ (2)
On equating (1) and (3) we get,
$\text{ }\left( \dfrac{1}{4} \right)={{\left( \dfrac{1}{2} \right)}^{\text{m}}}\text{ }\Rightarrow \text{ (0}\text{.25) = (0}\text{.5}{{\text{)}}^{\text{m}}}\text{ }$
Take a natural log on LHS and RHs .We have,
$\begin{align}
  & \text{ ln (0}\text{.25) = m ln (0}\text{.5)} \\
 & \Rightarrow \text{ m = }\dfrac{\text{ln (0}\text{.25)}}{\text{ln (0}\text{.5)}}=\text{ }\dfrac{-1.386}{-0.693}=2 \\
 & \therefore \text{m}=2 \\
\end{align}$
Now , Lets first consider the rate constant for set 1 and set 2.Divide the rate constant for the set 1 by the set 2.We have ,
$\text{ }\dfrac{{{\text{(Rate)}}_{\text{1}}}}{{{\text{(Rate)}}_{\text{2}}}}\text{ = }\dfrac{\text{K }{{\left[ {{\text{A}}_{\text{1}}} \right]}^{\text{m}}}{{\left[ {{\text{B}}_{1}} \right]}^{\text{n}}}\text{ }}{\text{K }{{\left[ {{\text{A}}_{2}} \right]}^{\text{m}}}{{\left[ {{\text{B}}_{2}} \right]}^{\text{n}}}\text{ }}$
Let’s substitute the values from the table. We have,
$\text{ }\dfrac{{{\text{(Rate)}}_{\text{1}}}}{{{\text{(Rate)}}_{\text{2}}}}\text{ = }\dfrac{\text{ }{{\left[ {{\text{A}}_{1}} \right]}^{\text{m}}}{{\left[ {{\text{B}}_{1}} \right]}^{\text{n}}}\text{ }}{\text{ }{{\left[ {{\text{A}}_{2}} \right]}^{\text{m}}}{{\left[ {{\text{B}}_{2}} \right]}^{\text{n}}}\text{ }}\text{= }\dfrac{{{\left( \text{2}\text{.5 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}} \right)}^{\text{m}}}{{\left( \text{3}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}} \right)}^{\text{n}}}}{{{\left( \text{5}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}} \right)}^{\text{m}}}{{\left( \text{6}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}} \right)}^{\text{n}}}}\text{ }$ (1)
The ratio of rate constant for set 1 and set 2 is equal to,
$\text{ }\dfrac{{{\text{(Rate)}}_{\text{1}}}}{{{\text{(Rate)}}_{\text{2}}}}\text{ = }\dfrac{\left( 5.0\times {{10}^{-4}} \right)}{\left( 4.0\times {{10}^{-3}} \right)}\text{ = }\left( \dfrac{1}{8} \right)\text{ }$ (2)
We know that, the order of reaction with respect to A is 2.Thus, equation (1) becomes,
$\begin{align}
  & \text{ }\dfrac{1}{8}\text{= }\dfrac{{{\left( \text{2}\text{.5 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}} \right)}^{\text{m}}}{{\left( \text{3}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}} \right)}^{\text{n}}}}{{{\left( \text{5}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}} \right)}^{\text{m}}}{{\left( \text{6}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}} \right)}^{\text{n}}}}\text{ } \\
 & \Rightarrow \dfrac{1}{8}\text{= }\dfrac{{{\left( \text{2}\text{.5 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}} \right)}^{2}}{{\left( \text{3}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}} \right)}^{\text{n}}}}{{{\left( \text{5}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}} \right)}^{2}}{{\left( \text{6}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}} \right)}^{\text{n}}}} \\
 & \Rightarrow \dfrac{\text{2}\text{.5 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-7}}}}{\text{5}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-7}}}}\text{= }\dfrac{{{\left( \text{3}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}} \right)}^{\text{n}}}}{{{\left( \text{6}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}} \right)}^{\text{n}}}} \\
 & \Rightarrow \left( \dfrac{1}{2} \right)=\text{ }{{\left( \dfrac{1}{2} \right)}^{\text{n}}} \\
\end{align}$
Take a natural log on LHS and RHs .We have,
$\begin{align}
  & \text{ ln (0}\text{.5) = n ln (0}\text{.5)} \\
 & \Rightarrow \text{ n = }\dfrac{\text{ln (0}\text{.5)}}{\text{ln (0}\text{.5)}}=\text{ 1} \\
 & \therefore \text{n}=1 \\
\end{align}$
Thus, the order of reaction with respect to B is 1.
Part B) We have to find the rate constant for the reaction at $\text{ 300K }$ .The rate of reaction is equal to the concentration of the reactant A and B .The rate equation is given as ,
$\text{ Rate = K }{{\left[ \text{A} \right]}^{2}}{{\left[ \text{B} \right]}^{1}}\text{ }$ (4)
Let’s consider the rate of reaction with respect to set 1.Substitute the values form the set (1) in to the equation (4) we have,
$\begin{align}
  & \text{ }\left( 5.0\times {{10}^{-4}} \right)\text{ = K }{{\left[ 2.5\times {{10}^{-4}} \right]}^{2}}{{\left[ 3.0\times {{10}^{-5}} \right]}^{1}} \\
 & \Rightarrow \text{K = }\dfrac{\left( 5.0\times {{10}^{-4}} \right)}{\left( 6.25\times {{10}^{-8}} \right)\left( 3.0\times {{10}^{-5}} \right)} \\
 & \Rightarrow \text{K = }\dfrac{\left( 5.0\times {{10}^{-4}} \right)}{\left( 6.25\times {{10}^{-8}} \right)\left( 3.0\times {{10}^{-5}} \right)} \\
 & \therefore \text{K = 2}\text{.66 }\times {{10}^{8}}\text{ }{{\text{L}}^{\text{2}}}\text{mo}{{\text{l}}^{-2}}{{\text{s}}^{-1}}\text{ } \\
\end{align}$
Thus, rate of reaction at $\text{ 300K }$is equal to the$\text{ 2}\text{.66 }\times {{10}^{8}}\text{ }{{\text{L}}^{\text{2}}}\text{mo}{{\text{l}}^{-2}}{{\text{s}}^{-1}}\text{ }$.

Note:
The order of the reaction can be determined by the other methods. The methods are as:
 The half –life method: The half-life period $\text{ }{{\text{t}}_{{}^{1}/{}_{2}}}\text{ }$ is used to determine the nth –order of a reaction. The relation is,
$\text{ n = 1 + }\dfrac{\ln {{{\left( \text{ }{{\text{t}}_{{}^{1}/{}_{2}}}\text{ } \right)}_{1}}}/{{{\left( \text{ }{{\text{t}}_{{}^{1}/{}_{2}}}\text{ } \right)}_{2}}}\;}{\ln {{{a}_{2}}}/{{{a}_{1}}}\;}\text{ }$.