ext{The experimental data for the reaction,}\
2 A+B_{2} \longrightarrow 2 A B ext{ is}
\
\begi

Question

ext{The experimental data for the reaction,}\
2 A+B_{2} \longrightarrow 2 A B 	ext{ is}
\
\begin{array}{cccl}
	ext{Experiment} & [A] & [B_{2}] & 	ext{Rate } (	ext{mole s}^{-1}) \
1 & 0.50 & 0.50 & 1.6 	imes 10^{-4} \
2 & 0.50 & 1.00 & 3.2 	imes 10^{-4} \
3 & 1.00 & 1.00 & 3.2 	imes 10^{-4} \
\end{array}
\
	ext{The rate equation for the above data is}\

Professor PC Thomas Classes · Answer

(c) : For the reaction, $2 A+B_{2} ightleftharpoons 2 A B$,

Rate $\propto[A]^{x}\left[B_{2}ight]^{y}$.\
On substituting the given data, we get

From experiment 1,

\begin{equation*}
1.6 	imes 10^{-4} \propto[0.50]^{x}[0.50]^{y} 	ag{i}
\end{equation*}

From experiment 2 ,

\begin{equation*}
3.2 	imes 10^{-4} \propto[0.50]^{x}[1.00]^{y} 	ag{ii}
\end{equation*}

From experiment 3 ,

\begin{equation*}
3.2 	imes 10^{-4} \propto[1.00]^{x}[1.00]^{y} 	ag{iii}
\end{equation*}

On dividing equation (iii) by (ii), we get ,\
$1=\left[\frac{1.00}{0.50}ight]^{x} \Rightarrow 1=2^{x} \Rightarrow 2^{0}=2^{x} \Rightarrow x=0$\
Now, divide equation (ii) by equation (i) we get,\
$2=\left[\frac{1.00}{0.50}ight]^{y} \Rightarrow 2=2^{y} \Rightarrow y=1$\
Thus, rate equation is : Rate $=k[A]^{0}\left[B_{2}ight]^{1}=k\left[B_{2}ight]$\