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
ro/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.