Question

In a reaction between A and B the initial rate of reaction $\left(r_0\right)$ was measured for different initial concentrations of A and B as given below.
$\begin{array}{cccc}A/mol{L}^{-1}& 0.20& 0.20& 0.40\\ B/mol{L}^{-1}& 0.30& 0.10& 0.05\\ x_0/mol{L}^{-1}{s}^{-1}& 5.07\times {10}^{-5}& 5.07\times {10}^{-5}& 1.43\times {10}^{-4}\end{array}$
What is the order to the reaction with respect to A and B?

Answer 1

Rate law states that
Rate $=k\left[A{]}^x\right[B{]}^y$
$(Rate{)}_1=k[0.20{]}^x[0.30{]}^y=5.07\times {10}^{-5}\dots \dots (i\left)\right(Rate{)}_2=k\left[0.20{]}^x\right[0.10{]}^y=5.07\times {10}^{-5}\dots \dots \left(ii\right)(Rate{)}_3=k[0.40{]}^x[0.05{]}^y=1.43\times {10}^{-4}\dots \dots (iii)$
Dividing equation (i) by equation (ii)
$\frac{(Rate{)}_1}{(Rate{)}_2}=\frac{k\left[0.20{]}^x\right[0.30{]}^y}{k\left[0.20{]}^x\right[0.10{]}^y}=\frac{5.07\times {10}^{-5}}{5.07\times {10}^{-5}}=1[3{]}^y=[3{]}^0=1;y=0$
Now dividing equation (ii) by equation (iii)
$\frac{(Rate{)}_2}{(Rate{)}_3}=\frac{k\left[0.20{]}^x\right[0.10{]}^0}{k\left[0.40{]}^x\right[0.05{]}^0}=\frac{5.07\times {10}^{-5}}{1.43\times {10}^{-4}}{\left[\frac{1}{2}\right]}^x=\frac{1}{2.82}$
$2^x=2.82$
$xlog2=log2.82\therefore x=1.4957\simeq 1.5$
Order with respect to A = 1.5
Order with respect to B = 0