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
N/ mol L-1 - 0.30 - 0.10 - 0.05
r0/ mol L-1 S-1 - 5.07 x 10-5 - 5.07 x 10-5 - 1.43 x 10-4
What is the order of the reaction with respect to A and B?
Let the order of the reaction with respect to A be x and with respect to B be y.
Therefore,
r0 = k[A]x [B]y
5.07 x 10-5 = k[0.20]x[0.30]y ....(i)
5.07 x 10-5 = k[0.20]x[0.10]y ....(ii)
1.43 x 10-4 = k[0.40]x[0.05]y ....(iii)
Dividing equation (i) by (ii), we obtain
\(\frac{5.07 \times 10^{-5}}{5.07 \times 10^{-5}}\) = \(\frac{k[0.20]^x[0.30]^y}{k[0.20]^x[0.10]^y}\)
⇒ 1 = \(\frac{[0.30]^y}{[0.10]^y}\)
⇒ \(\Big(\frac{0.30}{0.10}\Big)^0\) = \(\Big(\frac{0.30}{0.10}\Big)^y\)
⇒ y = 0
Dividing equation (iii) by (ii), we obtain
\(\frac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}}\) = \(\frac{k[0.40]^x[0.05]^y}{k[0.20]^x[0.30]^y}\)
⇒ \(\frac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}}\) = \(\frac{[0.40]^x}{[0.20]^x}\) [Since y = 0, [0.05]y = [0.30]y = 1]
⇒ 2.821 = 2x
⇒ log 2.821 = x log 2 (Taking log on both sides)
⇒ x = \(\frac{log 2.821}{log 2}\)
= 1.496
= 1.5 (approximately)
Hence, the order of the reaction with respect to A is 1.5 and with respect to B is zero.