JEE Advance - Chemistry (2004 - No. 1)
For the given reactions, A + B $$\to$$ Products, following data were obtained
(a) Write the rate law expression
(b) Find the rate constant
| [Ao] | [Bo] | Ro (mol L-1 s-1) | |
|---|---|---|---|
| 1 | 0.1 | 0.1 | 0.05 |
| 2 | 0.2 | 0.1 | 0.10 |
| 3 | 0.1 | 0.2 | 0.05 |
(b) Find the rate constant
R₀ = k [A₀][B₀]; k = 0.5 sec⁻¹
R₀ = k [A₀]; k = 0.05 sec⁻¹
R₀ = k [B₀]; k = 0.5 sec⁻¹
R₀ = k [A₀]; k = 0.5 sec⁻¹
R₀ = k [A₀]²; k = 0.5 sec⁻¹
Explanation
According to the given data, we see that when the concentration of A alone is doubled (0.1 $$\to$$ 0.2 mol L$$-$$1), the rate of the reaction gets doubled (0.05 $$\to$$ 0.1 mol L$$-$$1 s$$-$$1). Therefore, the order with respect to A is 1.
When the concentration of B alone is doubled (0.1 $$\to$$ 0.2 mol L$$-$$1), the rate of the reaction remains unaltered. Therefore, the order with respect to B is 0.
(1) The rate equation is, Rate = $$k[A]{[B]^0} = k[A]$$
(2) Rate $$ = k[A]$$ or, $$k = {{rate} \over {[A]}}$$ or, $$k = {{0.05} \over {0.1}}$$ $$\therefore$$ k = 0.5 s$$-$$1
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