Derive the rate law equation for a third-order reaction A+2B→P. Express the integrated rate law in concentration terms.
The Correct Answer and Explanation is:
Rate Law and Integrated Rate Law for a Third-Order Reaction: A + 2B → P
Rate Law (Differential Form):
For the reaction:A+2B→PA + 2B \rightarrow PA+2B→P
The overall reaction is third-order (first-order in A and second-order in B). The rate law is expressed as:Rate=−d[A]dt=k[A][B]2\text{Rate} = -\frac{d[A]}{dt} = k[A][B]^2Rate=−dtd[A]=k[A][B]2
Integrated Rate Law (in terms of concentration):
To derive the integrated rate law, we must account for the stoichiometry. Let:
- [A]0[A]_0[A]0 and [B]0[B]_0[B]0 be the initial concentrations of A and B, respectively.
- Let xxx be the amount of A reacted at time ttt. Then, the change in concentration of A is [A]0−x[A]_0 – x[A]0−x, and since 2 moles of B react with 1 mole of A, the concentration of B is [B]0−2x[B]_0 – 2x[B]0−2x.
Substituting into the rate law:dxdt=k([A]0−x)([B]0−2x)2\frac{dx}{dt} = k([A]_0 – x)([B]_0 – 2x)^2dtdx=k([A]0−x)([B]0−2x)2
This is a complex differential equation and typically cannot be integrated analytically in a simple form unless specific assumptions are made. However, if [B]_0 ≫ [A]_0, then [B] remains approximately constant during the reaction. This is known as the pseudo-first-order approximation.
Under this approximation:Rate=k′[A],where k′=k[B]2\text{Rate} = k'[A], \quad \text{where } k’ = k[B]^2Rate=k′[A],where k′=k[B]2
Integrating:d[A][A]=−k′[B]2dt⇒ln[A]=−k[B]2t+ln[A]0⇒[A]=[A]0e−k[B]2t\frac{d[A]}{[A]} = -k'[B]^2 dt \Rightarrow \ln[A] = -k[B]^2t + \ln[A]_0 \Rightarrow [A] = [A]_0 e^{-k[B]^2t}[A]d[A]=−k′[B]2dt⇒ln[A]=−k[B]2t+ln[A]0⇒[A]=[A]0e−k[B]2t
This is valid under pseudo-first-order conditions only.
Full Explanation
In chemical kinetics, the rate law expresses the speed of a reaction as a function of the concentrations of reactants. For a reaction A+2B→PA + 2B \rightarrow PA+2B→P, it is third-order overall: first-order in A and second-order in B. The rate law is written as:Rate=−d[A]dt=k[A][B]2\text{Rate} = -\frac{d[A]}{dt} = k[A][B]^2Rate=−dtd[A]=k[A][B]2
Here, kkk is the rate constant, and the exponents correspond to the reaction order with respect to each reactant. The negative sign indicates that the concentration of A decreases over time.
To determine the integrated rate law, which relates concentrations to time, we consider the stoichiometric relationship between the reactants. If xxx is the amount of A reacted at time ttt, then [A]=[A]0−x[A] = [A]_0 – x[A]=[A]0−x, and [B]=[B]0−2x[B] = [B]_0 – 2x[B]=[B]0−2x, because two moles of B react per mole of A. Substituting these into the rate law yields:dxdt=k([A]0−x)([B]0−2x)2\frac{dx}{dt} = k([A]_0 – x)([B]_0 – 2x)^2dtdx=k([A]0−x)([B]0−2x)2
This equation is not easily integrable in closed form and often requires numerical methods or simplifying assumptions. A common simplification is the pseudo-first-order approximation, applied when [B]_0 is much larger than [A]_0. In this case, [B] remains nearly constant during the reaction, and the rate law simplifies to:Rate=k′[A],with k′=k[B]2\text{Rate} = k’ [A], \quad \text{with } k’ = k[B]^2Rate=k′[A],with k′=k[B]2
This results in a first-order integrated rate law:[A]=[A]0e−k[B]2t[A] = [A]_0 e^{-k[B]^2 t}[A]=[A]0e−k[B]2t
This exponential decay equation describes how [A] decreases over time in such conditions.
