Question

Consider a reaction: 2A (aq) + 3B (aq) → C

The rate law is rate = k [A] [B]with rate constant k = ln($\frac{9}{4}$) L mol min$^{-1}$. An experiment started with 2 M of A and 6 M of B. Find the half lifetime of 'A' in seconds under given conditions.

Answer 1

5

Answer 2

The reaction is $2A(aq) + 3B(aq) \rightarrow C$. The rate law is rate = $k[A][B]$. The rate of disappearance of A is given by $-\frac{1}{2}\frac{d[A]}{dt} = k[A][B]$, so $\frac{d[A]}{dt} = -2k[A][B]$. The initial concentrations are $[A]_0 = 2$ M and $[B]_0 = 6$ M. From the stoichiometry, when $x$ moles of A react, $\frac{3}{2}x$ moles of B react. Let $[A]$ and $[B]$ be the concentrations at time $t$. The decrease in concentration of A is $[A]_0 - [A]$, and the decrease in concentration of B is $[B]_0 - [B]$. So, $[B]_0 - [B] = \frac{3}{2}([A]_0 - [A])$. Substituting the initial concentrations: $6 - [B] = \frac{3}{2}(2 - [A])$. $6 - [B] = 3 - \frac{3}{2}[A]$. $[B] = 6 - 3 + \frac{3}{2}[A] = 3 + \frac{3}{2}[A]$.

Substitute this expression for $[B]$ into the rate equation: $\frac{d[A]}{dt} = -2k[A](3 + \frac{3}{2}[A]) = -2k[A] \cdot \frac{3}{2}(2 + [A]) = -3k[A](2 + [A])$. Separate the variables: $\frac{d[A]}{[A](2 + [A])} = -3k dt$.

To find the half-life of A, $t_{1/2}$, we need to integrate from $t=0$ to $t=t_{1/2}$, and from $[A]=[A]_0=2$ to $[A]=[A]_0/2=1$. $\int_{2}^{1} \frac{d[A]}{[A](2 + [A])} = \int_{0}^{t_{1/2}} -3k dt$.

Using partial fractions, $\frac{1}{[A](2 + [A])} = \frac{1}{2[A]} - \frac{1}{2(2 + [A])}$. So, the integral becomes: $\int_{2}^{1} \frac{1}{2} (\frac{1}{[A]} - \frac{1}{2 + [A]}) d[A] = -3k \int_{0}^{t_{1/2}} dt$. $\frac{1}{2} [\ln|[A]| - \ln|2 + [A]|]_{2}^{1} = -3k [t]_{0}^{t_{1/2}}$. $\frac{1}{2} [\ln|\frac{[A]}{2 + [A]}|]_{2}^{1} = -3k t_{1/2}$. $\frac{1}{2} (\ln|\frac{1}{2 + 1}| - \ln|\frac{2}{2 + 2}|) = -3k t_{1/2}$. $\frac{1}{2} (\ln|\frac{1}{3}| - \ln|\frac{2}{4}|) = -3k t_{1/2}$. $\frac{1}{2} (\ln(\frac{1}{3}) - \ln(\frac{1}{2})) = -3k t_{1/2}$. $\frac{1}{2} \ln(\frac{1/3}{1/2}) = -3k t_{1/2}$. $\frac{1}{2} \ln(\frac{2}{3}) = -3k t_{1/2}$. $\ln(\frac{2}{3}) = -6k t_{1/2}$. $-\ln(\frac{3}{2}) = -6k t_{1/2}$. $\ln(\frac{3}{2}) = 6k t_{1/2}$. $t_{1/2} = \frac{\ln(\frac{3}{2})}{6k}$.

Given $k = \ln(\frac{9}{4})$ L mol$^{-1}$ min$^{-1}$. We can write $\ln(\frac{9}{4}) = \ln((\frac{3}{2})^2) = 2 \ln(\frac{3}{2})$. Substitute the value of $k$: $t_{1/2} = \frac{\ln(\frac{3}{2})}{6 \cdot 2 \ln(\frac{3}{2})} = \frac{\ln(\frac{3}{2})}{12 \ln(\frac{3}{2})} = \frac{1}{12}$ minutes.

The question asks for the half-life in seconds. $t_{1/2} = \frac{1}{12}$ min $\times \frac{60 \text{ s}}{1 \text{ min}} = 5$ seconds.

The final answer is $\boxed{5}$.

Explanation of the solution:

1. Write the rate equation for the disappearance of A based on the given rate law and stoichiometry.
2. Establish a relationship between the concentrations of A and B at any time $t$ using the initial concentrations and stoichiometry.
3. Substitute the relationship between $[A]$ and $[B]$ into the rate equation to get a differential equation involving only $[A]$ and $t$.
4. Solve the differential equation by separating variables and integrating. Use partial fractions for the integration of the concentration term.
5. Apply the limits of integration: from $t=0$ to $t=t_{1/2}$, and from $[A]=[A]_0$ to $[A]=[A]_0/2$.
6. Substitute the given initial concentration of A and the value of the rate constant $k$ to calculate the half-life of A in minutes.
7. Convert the half-life from minutes to seconds.

The final answer is $\boxed{5}$.