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Question: $A \xrightarrow{k_1 = ln 9 min^{-1}} B$ and $A \xrightarrow{k_2 = ln 9 min^{-1}} C$, the time (in s...

A→k1=ln9min−1BA \xrightarrow{k_1 = ln 9 min^{-1}} B and A→k2=ln9min−1CA \xrightarrow{k_2 = ln 9 min^{-1}} C, the time (in second) when concentration of A, B, C becomes equal is

Answer

15

Explanation

Solution

The given reactions are parallel first-order reactions: A→k1=ln⁡9 min−1BA \xrightarrow{k_1 = \ln 9 \text{ min}^{-1}} B A→k2=ln⁡9 min−1CA \xrightarrow{k_2 = \ln 9 \text{ min}^{-1}} C

Let [A]0[A]_0 be the initial concentration of A. We assume the initial concentrations of products B and C are zero.

The rate constants are given as k1=ln⁡9 min−1k_1 = \ln 9 \text{ min}^{-1} and k2=ln⁡9 min−1k_2 = \ln 9 \text{ min}^{-1}. Since k1=k2k_1 = k_2, the rates of formation of B and C are equal. Consequently, their concentrations will also be equal at any given time, i.e., [B]t=[C]t[B]_t = [C]_t.

We are asked to find the time (t) when the concentrations of A, B, and C become equal: [A]t=[B]t=[C]t[A]_t = [B]_t = [C]_t. Let this common concentration be XX. So, [A]t=X[A]_t = X, [B]t=X[B]_t = X, and [C]t=X[C]_t = X.

According to the conservation of mass for a first-order parallel reaction, the initial concentration of the reactant is equal to the sum of the concentrations of the reactant and products at any time t: [A]0=[A]t+[B]t+[C]t[A]_0 = [A]_t + [B]_t + [C]_t

Substitute the condition [A]t=[B]t=[C]t=X[A]_t = [B]_t = [C]_t = X: [A]0=X+X+X[A]_0 = X + X + X [A]0=3X[A]_0 = 3X Therefore, X=[A]03X = \frac{[A]_0}{3}.

So, we need to find the time t when [A]t=[A]03[A]_t = \frac{[A]_0}{3}.

For a first-order reaction, the concentration of the reactant A at time t is given by: [A]t=[A]0e−(k1+k2)t[A]_t = [A]_0 e^{-(k_1 + k_2)t}

Substitute [A]t=[A]03[A]_t = \frac{[A]_0}{3} into the equation: [A]03=[A]0e−(k1+k2)t\frac{[A]_0}{3} = [A]_0 e^{-(k_1 + k_2)t}

Divide both sides by [A]0[A]_0: 13=e−(k1+k2)t\frac{1}{3} = e^{-(k_1 + k_2)t}

Take the natural logarithm on both sides: ln⁡(13)=−(k1+k2)t\ln\left(\frac{1}{3}\right) = -(k_1 + k_2)t −ln⁡3=−(k1+k2)t-\ln 3 = -(k_1 + k_2)t ln⁡3=(k1+k2)t\ln 3 = (k_1 + k_2)t

Solve for t: t=ln⁡3k1+k2t = \frac{\ln 3}{k_1 + k_2}

Now, calculate the sum of the rate constants: k1+k2=ln⁡9 min−1+ln⁡9 min−1k_1 + k_2 = \ln 9 \text{ min}^{-1} + \ln 9 \text{ min}^{-1} k1+k2=2ln⁡9 min−1k_1 + k_2 = 2 \ln 9 \text{ min}^{-1}

We know that ln⁡9=ln⁡(32)=2ln⁡3\ln 9 = \ln (3^2) = 2 \ln 3. So, k1+k2=2(2ln⁡3)=4ln⁡3 min−1k_1 + k_2 = 2 (2 \ln 3) = 4 \ln 3 \text{ min}^{-1}.

Substitute this value into the equation for t: t=ln⁡34ln⁡3t = \frac{\ln 3}{4 \ln 3} t=14 mint = \frac{1}{4} \text{ min}

The question asks for the time in seconds. Convert minutes to seconds: t=14 min×60 s/mint = \frac{1}{4} \text{ min} \times 60 \text{ s/min} t=15 secondst = 15 \text{ seconds}

The final answer is 15\boxed{\text{15}}.

Explanation of the solution:

  1. Recognize the parallel first-order reactions with equal rate constants (k1=k2k_1 = k_2). This implies [B]t=[C]t[B]_t = [C]_t.
  2. Set the condition [A]t=[B]t=[C]t=X[A]_t = [B]_t = [C]_t = X.
  3. Use conservation of mass: [A]0=[A]t+[B]t+[C]t=3X[A]_0 = [A]_t + [B]_t + [C]_t = 3X, so X=[A]0/3X = [A]_0/3.
  4. Equate [A]t[A]_t to [A]0/3[A]_0/3: [A]0/3=[A]0e−(k1+k2)t[A]_0/3 = [A]_0 e^{-(k_1+k_2)t}.
  5. Simplify to 1/3=e−(k1+k2)t1/3 = e^{-(k_1+k_2)t}.
  6. Take natural logarithm: ln⁡(1/3)=−(k1+k2)t\ln(1/3) = -(k_1+k_2)t, which simplifies to ln⁡3=(k1+k2)t\ln 3 = (k_1+k_2)t.
  7. Solve for tt: t=ln⁡3k1+k2t = \frac{\ln 3}{k_1+k_2}.
  8. Substitute k1=k2=ln⁡9=2ln⁡3k_1 = k_2 = \ln 9 = 2\ln 3. So k1+k2=4ln⁡3k_1+k_2 = 4\ln 3.
  9. Calculate t=ln⁡34ln⁡3=14t = \frac{\ln 3}{4\ln 3} = \frac{1}{4} min.
  10. Convert minutes to seconds: t=14×60=15t = \frac{1}{4} \times 60 = 15 seconds.