Question

The following results have been obtained during the kinetic studies of the reaction, $2A+B\to C+D$
$\begin{array}{cccc}Experiment& \left[A\right]mol{L}^{-1}& \left[B\right]mol{L}^{-1}& Initialrateofformationof\\ & & & D/mol{L}^{-1}min\\ I& 0.1& 0.1& 6.0\times {10}^{-3}\\ II& 0.3& 0.2& 7.2\times {10}^{-2}\\ III& 0.3& 0.4& 2.88\times {10}^{-1}\\ IV& 0.4& 0.1& 2.40\times {10}^{-2}\end{array}$
Determine the rate law and the rate constant for the reaction.

Answer 1

Rate law may be expressed as
Rate $=k\left[A{]}^x\right[B{]}^y$
$(Rate{)}_1=6.0\times {10}^{-3}=k(0.1{)}^x(0.1{)}^y\dots \dots (i\left)\right(Rate{)}_2=7.2\times {10}^{-2}=k\left(0.3{)}^x\right(0.2{)}^y\dots \dots \left(ii\right)(Rate{)}_3=2.88\times {10}^{-1}=k(0.3{)}^x(0.4{)}^y\dots \dots (iii\left)\right(Rate{)}_4=2.40\times {10}^{-2}=k\left(0.4{)}^x\right(0.1{)}^y\dots \dots \left(iv\right)\frac{(Rate{)}_1}{(Rate{)}_4}=\frac{6.0\times {10}^{-3}}{2.40\times {10}^{-2}}=\frac{k\left(0.1{)}^x\right(0.1{)}^y}{k\left(0.4{)}^x\right(0.1{)}^y}$
or $\frac{1}{4}=\frac{(0.1{)}^x}{(0.4{)}^x}={\left(\frac{1}{4}\right)}^x$
$\therefore x=1\frac{(Rate{)}_2}{(Rate{)}_3}=\frac{7.2\times {10}^{-2}}{2.88\times {10}^{-1}}=\frac{k\left(0.3{)}^x\right(0.2{)}^y}{k\left(0.3{)}^x\right(0.4{)}^y}$
or $\frac{1}{4}=\frac{(0.2{)}^y}{(0.4{)}^y}={\left(\frac{1}{2}\right)}^y$
$\therefore y=2$
Rate law expression is given by
Rate $=k\left[A\right][B{]}^2$
Rate constant k can be determined by placing the values of A, B and rate of formation of D. By taking the values from experiment II.
Rate $=k\left[A\right][B{]}^{2}k=\frac{Rate}{\left[A\right][B{]}^2}=\frac{7.2\times {10}^{-2}mol{L}^{-1}mi{n}^{-1}}{\left(0.3mol{L}^{-1}\right)(0.2mol{L}^{-1}{)}^2}=6.0mo{l}^{-2L^{2}mi{n}^{-1}}k=6.0mo{l}^{-2}{L}^{2}mi{n}^{-1}$