The following results have been obtained during the kinetic studies of the reaction: 2A + B → C + D
Experiment - A/mol L-1 - B/ mol L-1 - Initial rate of formation of D/mol L-1 min-1
I - 0.1 - 0.1 - 6.0 x 10-3
II - 0.3 - 0.2 - 7.2 x 10-2
III - 0.3 - 0.4 - 2.88 x 10-1
IV - 0.4 - 0.1 - 2.40 x 10-2
Determine the rate law and the rate constant for the reaction.
Let the order of the reaction with respect to A be x and with respect to B be y.
Therefore, rate of the reaction is given by,
Rate = k[A]x [B]y
According to the question,
6.0 x 10-3 = k[0.1]x [0.1]y .....(i)
7.2 x 10-2 = k[0.3]x [0.2]y .....(ii)
2.88 x 10-1 = k[0.3]x [0.4]y .....(iii)
2.40 x 10-2 = k[0.4]x [0.1]y .....(iv)
Dividing equation (iv) by (i), we obtain
\(\frac{2.40 \times 10^{-2}}{6.0 \times 10^{-3}}\) = \(\frac{k[0.4]^x [0.1]^y}{k[0.1]^x [0.1]^y}\)
⇒ 4 = \(\frac{[0.4]^x}{[0.1]^x}\)
⇒ 4 = \(\Big(\frac{0.4}{0.1}\Big)^x\)
⇒ (4)1 = 4x
⇒ x = 1
Dividing equation (iii) by (ii), we obtain
Therefore, the rate law is Rate = k [A] [B]2
⇒ k = \(\frac{Rate}{[A][B]^2}\)
From experiment I, we obtain
k = \(\frac{6.0 \times 10^{-3} mol\; L^{-1}\; min^{-1}}{(0.1\; mol\;L^{-1})(0.1\; mol\;L^{-1})^2}\)
= 6.0 L2mol-2min-1
From experiment II, we obtain
k = \(\frac{7.2 \times 10^{-2} mol\; L^{-1}\; min^{-1}}{(0.3\; mol\;L^{-1})(0.4\; mol\;L^{-1})^2}\)
= 6.0 L2mol-2min-1
From experiment IV, we obtain
k = \(\frac{2.40 \times 10^{-2} mol\; L^{-1}\; min^{-1}}{(0.4\; mol\;L^{-1})(0.1\; mol\;L^{-1})^2}\)
= 6.0 L2mol-2min-1
Therefore, rate constant, k = 6.0 L2mol-2min-1