Question

Consider the hypothetical reaction $\mathit{A}\mathbf{+}\mathit{B}\mathbf{+}\mathbf{2}\mathit{C}\mathbf{\to }\mathbf{2}\mathit{D}\mathbf{+}\mathbf{3}\mathit{E}$where the rate law is

An experiment is carried out where${\mathbf{\left[}\mathbf{A}\mathbf{\right]}}_{\mathbf{0}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathit{M}$

,${\left[B\right]}_{\mathbf{0}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathit{M}$ and${\left[C\right]}_{\mathbf{0}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathit{M}$. The reaction is started and after seconds, the concentration of A is$\mathbf{3}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{M}$.

(a) Calculate the value offor this reaction.

(b) Calculate half-life for this experiment.

(c) Calculate concentration of A after 13.0seconds.

(d) Calculate the concentration of Cafter 13.0seconds.

Short answer

(a) The value offor this reaction is.

(b) The half-life for this experiment is5.77 seconds.

(c) The concentration ofafter13 secondsis.

(d) Calculate the concentration of C after 13 seconds is .

Step-by-step solution

Determine the value of the rate constant

Since the reaction is first order with respect to A, we can determine the value of the reaction rate constant K' with respect to A:

$\mathit{l}\mathit{n}\left[A\right]\mathbf{=}\mathbf{-}\mathit{K}\mathbf{\text{'}}\mathit{t}\mathbf{+}\mathit{l}\mathit{n}{\left[A\right]}_{\mathbf{0}}$

Determine the value for the rate constant as:

$\begin{array}{l}K\text{'}=\frac{\ln {\left[A\right]}_0-\ln \left[A\right]}{t}\\ K\text{'}=\frac{\ln \left[1.0\times {10}^{-2}\right]-\ln \left[3.8\times {10}^{-3}\right]}{8\sec }\\ K\text{'}=0.12{\mathrm{s}}^{-1}\end{array}$

Determine the overall reaction rate constant:

$\begin{array}{l}K\text{'}=K{\left[B\right]}^2\\ K=\frac{K\text{'}}{{\left[B\right]}^2}\\ K=\frac{0.12{\mathrm{s}}^{-1}}{{\left(3.0\mathrm{M}\right)}^2}\\ K=0.013{\mathrm{mol}}^2{\mathrm{L}}^{-2}{\mathrm{s}}^{-1}\end{array}$

Determine the half-life of the given experiment.

Since the reaction is first-order with respect to A:

$\begin{array}{c}t\left(\frac{1}{2}\right)=\frac{\ln 2}{K\text{'}}\\ =\frac{\ln 2}{0.12{\mathrm{s}}^{-1}}\\ =5.8\mathrm{s}\end{array}$

Calculation of concentration of  after  seconds using rate law equation

Consider the equation to determine the concentration as:

$\ln \left[A\right]=-K\text{'}t+\ln {\left[A\right]}_0$

Substitute the values and solve:

$\begin{array}{l}\ln \left[A\right]=0.12s^{-1}\times 135\times \ln \left(1.0\times {10}^{-2}\right)\\ \left[A\right]=2.10\times {10}^{-3}\mathrm{M}\end{array}$

Calculation of concentration of  after seconds.

The concentration of C is:

${\left[C\right]}_0-2\Delta \left[A\right]$

Solve further as:

$\begin{array}{c}\Delta \left[A\right]=1.0\times {10}^{-2}M-0.0021M\\ =0.008M\end{array}$

Consider the equation:

$\begin{array}{l}\left[C\right]={\left[C\right]}_0-2\Delta \left[A\right]\\ \left[C\right]=2.0M-2\left(0.008\mathrm{M}\right)\\ \left[C\right]=2.0\mathrm{M}\end{array}$