The initial rates of reaction 3A +2B + C → Products , at different in…

The correct answer is option 2. 2, 1, 0.

Let us suppose that the orders w.r.t. A, B, and C are \(\alpha \), \(\beta \), and \(\gamma \) respectively. Then

\(5.0 × 10^{-3} = (0.010)^{\alpha}(0.005)^{\beta}(0.010)^{\gamma}\) ---------(i)

\(5.0 × 10^{-3} = (0.010)^{\alpha}(0.005)^{\beta}(0.015)^{\gamma}\) ---------(ii)

\(1.0 × 10^{-2} = (0.010)^{\alpha}(0.010)^{\beta}(0.010)^{\gamma}\) ---------(iii)

\(1.25 × 10^{-3} = (0.005)^{\alpha}(0.005)^{\beta}(0.010)^{\gamma}\) ---------(iv)

Dividing (i) by (ii), we get

\(\frac{5.0 × 10^{-3}}{5.0 × 10^{-3}} = \frac{(0.010)^{\alpha}(0.005)^{\beta}(0.010)^{\gamma}}{(0.010)^{\alpha}(0.005)^{\beta}(0.015)^{\gamma}}\)

⇒ \(1 = \left(\frac{0.010}{0.015}\right)^{\gamma}\)

⇒\(\left(\frac{2}{3}\right)^{\gamma} = 1 = \left(\frac{2}{3}\right)^0\)

∴ \(\gamma = 0\)

Now, dividing (iii) by(ii), we get

\(\frac{1.0 × 10^{-2}}{5.0 × 10^{-3}} = \frac{(0.010)^{\alpha}(0.010)^{\beta}(0.010)^{\gamma}}{(0.010)^{\alpha}(0.005)^{\beta}(0.015)^{\gamma}}\)

⇒ \(2 = (2)^{\beta}\left(\frac{2}{3}\right)^{\gamma}\)

⇒ \(2 = (2)^{\beta}\left(\frac{2}{3}\right)^0\)       [Since, \(\gamma \) = 0]

⇒ \(2^1 = (2)^{\beta}\)

∴ \(\beta = 1\)

Also, dividing (i) by (iv), we get

\(\frac{5.0 × 10^{-3}}{1.25 × 10^{-3}} = \frac{(0.010)^{\alpha}(0.005)^{\beta}(0.010)^{\gamma}}{(0.005)^{\alpha}(0.005)^{\beta}(0.010)^{\gamma}}\)

⇒ \(4 = (2)^{\alpha}\)

⇒ \(2^2 = (2)^{\alpha}\)

∴ \(\alpha = 2\)

The orders with respect to the reactants A, B, and C are 2, 1,and 0 respectively