Question

In a reaction between A and B the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below:

A/mol L -1	0.20	0.20	0.40
B/mol L -1	0.30	0.10	0.05
r o/mol L-1 s-1	5.07x10-5	5.07x10-5	1.43x10-4

What is the order of the reaction with respect to A and B?

Answer 1

Let the order of the reaction with respect to A be x and with respect to B be y.
Therefore,
$r_0 = k[A]^x[B]^y$

$5.07\times 10^{-5} = k [0.20]^x[0.30]^y$ $......(1)$

$5.07\times 10^{-5} = k [0.20]^x[0.10]^y$ $......(2)$

$1.43\times 10^{-4} = k [0.40]^x[0.05]^y$ $......(3)$

Dividing equation (i) by (ii), we obtain

$\frac {5.07\times10^{-5} }{5.07\times10^{-5} }$= $\frac {k [0.20]^x[0.30]^y}{k [0.20]^x[0.30]^y}$

$1 = \frac {[0.30]^y}{[0.10]^y}$

$(\frac {0.30}{0.10})^0 = (\frac {0.30}{0.10})^y$

$y = 0$

Dividing equation (iii) by (i), we obtain

$\frac {1.43\times10^{-4}}{5.07\times 10^{-5}} = \frac {k [0.40]^x[0.05]^y}{k [0.20]^x[0.30]^y}$

$\frac {1.43\times10^{-4}}{5.07\times 10^{-5}}$ = $\frac {[0.40]^x}{[0.20]^x}$ $(since \ y = 0)$

$2.821 = 2x$

$log\ 2.821 = x log \ 2$ (taking log on both side)

$x = \frac {log\ 2.821}{log\ 2 }$

$x = 1.496$

$x = 1.5 \ (approximately)$

Hence, the order of the reaction with respect to A is 1.5 and with respect to B is zero.