Question

At 25°C in CCl₄ solvent, the reaction ${\text{I+I→I}}_{\text{2}}$ is second order in the concentration of the iodine atoms. The rate constant k has been measured as 8.2 × 10⁹ L mol^-1. Suppose the initial concentration of I atoms is 1.00 × 10^-4 M. Calculate their concentration after 2.0 × 10^-6 s.

Short answer

The final concendration of iodine after$2.0\times {10}^{-6}s$is $2.3\times {10}^{-5}\text{M}$

Step-by-step solution

Rate expression:

For the seacond order reaction the rate equation is given as follows

$\frac{1}{c_t}=\frac{1}{c0}+2kt.........\left(1\right)$

Where C and C₀ is the initial and final concendrations of the reaction,k is the rate constant and t is the time in seaconds.

Calculation of final concendration:

$$" width="9" height="19" role="math">$$\begin{gathered} \text{Fromthegivendata} \\ {C}_0=1.00\times {10}^{-4}M \\ {C}_t\hspace{0.17em}=2.00\times {10}^{-6}s \\ k=8.2\times {10}^9{\text{Lmol}}^{-1} \\ \text{Substitutingthesevaluesinequation}-1 \\ \frac{1}{c_t}=\frac{1}{c0}+2kt \\ \frac{1}{c_t}=\frac{1}{1.00\times {10}^{-4}{\text{molL}}^{-1}}+2\times \left(8.2\times {10}^9{\text{molL}}^{-1}{\text{s}}^{-1}\right) \\ \frac{1}{c_t}=\left[\frac{1}{1.00\times {10}^{-4}{\text{molL}}^{-1}}+32.8\times {10}^3\right]{\text{molL}}^{-1} \\ \frac{1}{c_t}=4.28\times {10}^4{\text{molL}}^{-1} \\ {c}_t=\frac{1}{4.28\times {10}^4{\text{molL}}^{-1}}(Since1\text{M}={\text{molL}}^{-1} \\ {c}_t\hspace{0.17em}=\hspace{0.17em}2.3\times {10}^{-5}\text{M} \end{gathered}$$

Therefore the final concendration of iodine after$2.0\times {10}^{-6}$s is$2.3\times {10}^{-5}\text{M}$.