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Question: Consider the following reversible reaction. D $\rightleftharpoons$ L. The reaction is started with o...

Consider the following reversible reaction. D ⇌\rightleftharpoons L. The reaction is started with only D.

Both the forward and backward reactions follow first order kinetics.

If the half life of forward reaction is 8 hrs and that of backward reaction is 4 hrs then find the ratio of [L]/[D] at t = 8 hrs

Answer

7/17

Explanation

Solution

The problem involves a reversible first-order reaction D ⇌\rightleftharpoons L, starting with only D. We are given the half-lives of the forward and backward reactions and need to find the ratio of concentrations [L]/[D] at a specific time t = 8 hours.

1. Calculate the rate constants: For a first-order reaction, the half-life (t1/2t_{1/2}) is related to the rate constant (kk) by the formula: t1/2=ln⁡2kt_{1/2} = \frac{\ln 2}{k}

Given: Half-life of forward reaction, t1/2,f=8t_{1/2,f} = 8 hrs Half-life of backward reaction, t1/2,b=4t_{1/2,b} = 4 hrs

So, the forward rate constant, kf=ln⁡28 hr−1k_f = \frac{\ln 2}{8} \text{ hr}^{-1} And the backward rate constant, kb=ln⁡24 hr−1k_b = \frac{\ln 2}{4} \text{ hr}^{-1}

2. Determine equilibrium concentrations: At equilibrium, the rates of the forward and backward reactions are equal: kf[D]eq=kb[L]eqk_f[D]_{eq} = k_b[L]_{eq} Therefore, the ratio of equilibrium concentrations is: [L]eq[D]eq=kfkb=ln⁡2/8ln⁡2/4=48=12\frac{[L]_{eq}}{[D]_{eq}} = \frac{k_f}{k_b} = \frac{\ln 2 / 8}{\ln 2 / 4} = \frac{4}{8} = \frac{1}{2}

Let [D]0[D]_0 be the initial concentration of D. Since the reaction starts with only D, [L]0=0[L]_0 = 0. The total concentration remains constant: [D]t+[L]t=[D]0[D]_t + [L]_t = [D]_0. At equilibrium, [D]eq+[L]eq=[D]0[D]_{eq} + [L]_{eq} = [D]_0. Substitute [L]eq=12[D]eq[L]_{eq} = \frac{1}{2}[D]_{eq} into the conservation equation: [D]eq+12[D]eq=[D]0[D]_{eq} + \frac{1}{2}[D]_{eq} = [D]_0 32[D]eq=[D]0  ⟹  [D]eq=23[D]0\frac{3}{2}[D]_{eq} = [D]_0 \implies [D]_{eq} = \frac{2}{3}[D]_0 And [L]eq=12×23[D]0=13[D]0[L]_{eq} = \frac{1}{2} \times \frac{2}{3}[D]_0 = \frac{1}{3}[D]_0.

3. Use the integrated rate law for reversible first-order reactions: For a reversible first-order reaction D⇌LD \rightleftharpoons L starting with [L]0=0[L]_0 = 0, the concentration of L at time tt is given by: [L]t=[L]eq(1−e−(kf+kb)t)[L]_t = [L]_{eq}(1 - e^{-(k_f+k_b)t})

First, calculate the sum of rate constants: kf+kb=ln⁡28+ln⁡24=ln⁡2(18+28)=3ln⁡28k_f + k_b = \frac{\ln 2}{8} + \frac{\ln 2}{4} = \ln 2 \left(\frac{1}{8} + \frac{2}{8}\right) = \frac{3 \ln 2}{8}

Now, calculate the exponent term e−(kf+kb)te^{-(k_f+k_b)t} at t=8t = 8 hrs: −(kf+kb)t=−(3ln⁡28)×8=−3ln⁡2=−ln⁡(23)=−ln⁡8-(k_f+k_b)t = -\left(\frac{3 \ln 2}{8}\right) \times 8 = -3 \ln 2 = -\ln(2^3) = -\ln 8 So, e−(kf+kb)t=e−ln⁡8=18e^{-(k_f+k_b)t} = e^{-\ln 8} = \frac{1}{8}

Substitute this value into the equation for [L]t[L]_t: [L]t=13[D]0(1−18)[L]_t = \frac{1}{3}[D]_0 \left(1 - \frac{1}{8}\right) [L]t=13[D]0(78)=724[D]0[L]_t = \frac{1}{3}[D]_0 \left(\frac{7}{8}\right) = \frac{7}{24}[D]_0

4. Calculate [D] at t = 8 hrs: Using the conservation of mass: [D]t=[D]0−[L]t[D]_t = [D]_0 - [L]_t [D]t=[D]0−724[D]0=(1−724)[D]0=1724[D]0[D]_t = [D]_0 - \frac{7}{24}[D]_0 = \left(1 - \frac{7}{24}\right)[D]_0 = \frac{17}{24}[D]_0

5. Find the ratio [L]/[D] at t = 8 hrs: [L]t[D]t=724[D]01724[D]0=717\frac{[L]_t}{[D]_t} = \frac{\frac{7}{24}[D]_0}{\frac{17}{24}[D]_0} = \frac{7}{17}

The ratio of [L]/[D] at t = 8 hrs is 7/17.