Take the equation $$\ce{A(g) <=> B(g)}$$ where the reaction is first order with respect to A. The rate law is thus $$\frac{\mathrm d[\ce{A}]}{\mathrm dt}=-k[\ce{A}]$$ Say you run the reaction, starting with $x$ moles of A. Equilibrium is established at time $y$. If you run the reaction again, this time with $2x$ moles of A, won’t equilibrium be stablished at time $y$ as well? You double the rate of reaction, but you have twice as many molecules to react, so wouldn’t time remain the same? I’ve been told that the equilibrium effect dominates over the kinetic effect; that is, it will take a time that is greater than $y$ for the reaction to reach equilibrium. I cannot fathom the mathematical basis of such a conclusion. Can someone elucidate?

  • $\begingroup$ You've misunderstood that statement about the "equillibrium effect" (which is imo a rather stupid term). It only means that after some time, if there is a significant backward reaction, you cannot predict the state of the reaction from the kinetics of the forward reaction any more. $\endgroup$ – Karl May 3 '16 at 14:35

I you have an equilibrium with significant amounts of $\ce{A}$ and $\ce{B}$ you need to consider the backwards reaction. If it is first order in $\ce{B}$, the rate law is

$$\frac{\mathrm d[\mathrm{A}]}{\mathrm dt}= -k_1 \cdot [\mathrm{A}] + k_2 \cdot [\mathrm{B}] \tag{1}$$

If the initial concentration of $\ce{A}$ is $\mathrm{[A]}_0$ and the initial concentration of $\ce{B}$ is $0$ the solution of the differential equation is

$$\frac{\mathrm{[A]}}{\mathrm{[A]}_0} = \frac{1}{k_1 + k_2} \cdot \left(k_2 + k_1 \cdot \mathrm e^{-\left(k_1 + k_2\right) t}\right) \tag{2}$$

The equilibrium concentration of $\ce{A}$ is obtained as the limit for $t \to \infty$:

$$\frac{\mathrm{[A]_{eq}}}{\mathrm{[A]}_0} = \frac{k_2}{k_1 + k_2} \tag{3}$$

From $(2)$ and $(3)$ it follows that

$$\frac{\mathrm{[A]} - \mathrm{[A]_{eq}}}{\mathrm{[A]}_0} = k_1 \cdot \mathrm e^{-(k_1 + k_2) t} \tag{4}$$

or, logarithmically,

$$t = -\frac{1}{k_1 + k_2} \left(\ln\left(\mathrm{[A]} - \mathrm{[A]_{eq}}\right) - \ln\left(\mathrm{[A]}_0\right) - \ln\left(k_1\right)\right) \tag{6}$$

The criterion for equilibrium $\left|\mathrm{[A]} - \mathrm{[A]_{eq}}\right| \ll 1$ lets us neglect the term $\ln\left(\mathrm{[A]}_0\right)$ in equation $(6)$.

This shows that the time it takes to reach the equilibrium is independent of the initial concentration of $\ce{A}$.


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