# Effect of increasing product concentration on time of completion of reaction

This was the question asked in our chemistry exam:

In the experiment of determination of equilibrium constant of esterification of ethanol, if , instead of using clean dry bottle, if a bottle already containing slight amounts of ethanoate and water is used to carry out the reaction under identical (remaining) conditions, what would be the consequence?

According to our lecturer it's answer is "The reaction will take lesser time to reach equilibrium". But according to Le Chatelier's principle, increase in concentration of products shifts the reaction backwards. So it should take more time to reach equilibrium. Please help!!

In equilibrium reactions the equilibrium is approached exponentially in time. The lifetime of the exponential is the reciprocal of sum of the forwards and back rate constants multiplied by the equilibrium concentration of the species, thus for reaction $$\ce{A + B \leftrightharpoons C + D }$$ the lifetime of the reaction is $$\tau = 1/[k_1 (A_e+B_e) + k_{-1} (C_e+D_e) ]$$ where $$k_1$$ is the forward rate constant and $$k_{-1}$$ the backwards one, and subscript $$e$$ means equilibrium amount of a species. Thus if you add more material you change the equilibrium concentrations and so the lifetime gets shorter as its a reciprocal in the concentrations
The lifetime is the time taken to change to $$1/e$$ of the initial amount, approx to 38%, whereas the half-life is the time taken to halve the initial amount. The lifetime is the reciprocal of the total rate constant for reaction which is why it is used instead of the half-life.
(The exception is the simple $$\ce{A\leftrightharpoons B}$$ reaction when the lifetime is independent of the equilibrium amounts ). In equilibrium reactions the lifetime is also called the relaxation time.