Question

In unpolluted air at 300 K, the hydroxyl radical OHreacts with COwith a bimolecular rate constant of $\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{11}}\mathbf{}{\mathbf{Lmol}}^{\mathbf{-}\mathbf{1}}{\mathbf{s}}^{\mathbf{-}\mathbf{1}}$and with ${\mathbf{CH}}_{\mathbf{4}}$with a rate constant of$\mathbf{3}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\mathbf{}{\mathbf{Lmol}}^{\mathbf{-}\mathbf{1}}{\mathbf{s}}^{\mathbf{-}\mathbf{1}}$. Take the partial pressure of COin air to be constant atrole="math" localid="1663816781014" $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{27}}\mathbf{}\mathbf{atm}$and that of${\mathbf{CH}}_{\mathbf{4}}$to be$\mathbf{1}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{26}}\mathbf{}\mathbf{atm}$and assume that these are the primary mechanisms by which OHis consumed in the atmosphere. Calculate the half-life of OH under these conditions.

Short answer

The half-life of OH radicle under the given condition is 0.077 s.

Step-by-step solution

Definition of half-life

Half-Life is the time required for a quantity to reduce to half of its initial value.

Given information

The temperature in kelvin is 300 K, the rate constant of CO is$1.6\times {10}^{11}{\mathrm{Lmol}}^{-1}{\mathrm{s}}^{-1}$ and the rate constant of ${\mathrm{CH}}_4$isrole="math" localid="1663817101935" $3.8\times {10}^9{\mathrm{Lmol}}^{-1}{\mathrm{s}}^{-1}$.

Calculation of the half-life of the OH radicle.

The concentration is as follows

$\begin{array}{rcl}\left[\mathrm{CO}\right]& =& \frac{{\mathrm{P}}_{\mathrm{CO}}}{\mathrm{RT}}\\ \left[\mathrm{CO}\right]& =& 4\times {10}^{-11}{\mathrm{molL}}^{-1}\\ \left[{\mathrm{CH}}_4\right]& =& \frac{{\mathrm{P}}_{{\mathrm{CH}}_4}}{\mathrm{RT}}\\ \left[{\mathrm{CH}}_4\right]& =& 6.8\times {10}^{-10}{\mathrm{molL}}^{-1}\end{array}$

Now, calculate the role="math" localid="1663817401261" $\frac{\mathrm{d}\left[\mathrm{OH}\right]}{\mathrm{dt}}$as follows:

role="math" localid="1663817658707" $\begin{array}{rcl}\frac{\mathrm{d}\left[\mathrm{OH}\right]}{\mathrm{dt}}& =& -{\mathrm{k}}_{\mathrm{CO}}\left[\mathrm{CO}\right]\left[\mathrm{OH}\right]-{\mathrm{k}}_{{\mathrm{CH}}_4}\left[{\mathrm{CH}}_4\right]\left[\mathrm{OH}\right]\\ & =& -\left(1.6\times {10}^{11}{\mathrm{Lmol}}^{-1}{\mathrm{s}}^{-1}\right)\left(4\times {10}^{-11}{\mathrm{Lmol}}^{-1}{\mathrm{s}}^{-1}\right)\left[\mathrm{OH}\right]-\left(3.8\times {10}^9{\mathrm{Lmol}}^{-1}{\mathrm{s}}^{-1}\right)\left(6.8\times {10}^{-10}{\mathrm{Lmol}}^{-1}{\mathrm{s}}^{-1}\right)\left[\mathrm{OH}\right]\\ & =& -9{\mathrm{s}}^{-1}\left[\mathrm{OH}\right]\\ & =& -{\mathrm{k}}_{\mathrm{eff}}\left[\mathrm{OH}\right]\end{array}$

The effective rate constant $k_{\mathrm{eff}}$is $9{\mathrm{s}}^{-1}$. The calculation of half-life is which is as follows:

$\begin{array}{rcl}{t}_{1/2}& =& \frac{\ln {\displaystyle }{\displaystyle 2}}{k}\\ & =& \frac{0.6931}{k}\\ & =& \frac{0.6931}{9.0}\\ & =& 0.077\mathrm{s}\end{array}$