Practicing Success
The initial rates of reaction \(3A +2B + C → Products\), at different initial concentrations are given below:
The order with respect to the reactants A, B, and C are respectively: |
3, 2, 1 2, 1, 0 3, 2, 0 2, 2, 0 |
2, 1, 0 |
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 |