Question

HO₂ is a highly reactive chemical species that plays a role in atmospheric chemistry. The rate of the gas-phase reaction

${\mathbf{\text{HO}}}_{\mathbf{2}}\left(g\right)\mathbf{}\mathbf{+}{\mathbf{\text{HO}}}_{\mathbf{2}}\left(g\right){\mathbf{\text{→H}}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{2}}\left(g\right)$

is second order in [HO₂], with a rate constant at 25°C of 1.4 × 10⁹ L mol^-1s^-1. Suppose some HO₂ with an initial concentration of 2.0 × 10^-8 M could be confined at 25°C. Calculate the concentration that would remain after 1.0 s, assuming no other reactions take place.

Short answer

The concentration ofH₂O was found to be$3.51\times {10}^{-10}\text{M}$

Step-by-step solution

Step-1:  Relation between reactants and products :

In a second -order reaction rate of reaction is directly proportional to the square of the concentration of reactants, so the expression is given as

$\begin{array}{l}\text{2A→products}\\ \text{Rater=K}{\left[A\right]}^2........\left(1\right)\end{array}$

Where k is the rate constant A is the concentration of reactants.

Step-2: Rate expression:

The rate law for second order reaction is given as

$\begin{array}{l}\frac{1}{\left[C_0\right]}-\frac{1}{\left[C_t\right]}=kt..........\left(2\right)\\ \text{Where}\left[C_0\right]=\text{initialconcendrationofreactents}\\ \left[C_t\right]=\text{finalconcendrationofreactents}\\ K=\text{rateconstant}\\ t=\text{timeinseaconds}\\ \text{Fromthegivendatesubstitutingthevaluesgiveninthequestionintheequation1}\\ C_0=2.0\times {10}^{-8}M\\ C_t=?\\ K=1.4\times {10}^9{\text{Lmol}}^{-1}{s}^{-1}\\ t=10\text{seaconds}\\ \frac{1}{\left[C_t\right]}=\frac{1}{2.0\times {10}^{-8}mol{L}^{-1}}+2\times \left(1.4\times {10}^9{\text{molL}}^{-1}{s}^{-1}\right)\times \left(1.0\text{s}\right)\\ \frac{1}{\left[C_t\right]}=\left(\frac{1}{2.0\times {10}^{-8}{\text{molL}}^{-1}}+2.8\times {10}^9\right){\text{molL}}^{-1}\\ \frac{1}{\left[C_t\right]}=\frac{57}{2.0\times {10}^{-8}}{\text{molL}}^{-1}\\ \frac{1}{\left[C_t\right]}=28.5\times {10}^8{\text{molL}}^{-1}\\ \frac{1}{\left[C_t\right]}=\frac{1}{28.5\times {10}^8}{\text{molL}}^{-1}\\ \left[C_t\right]=0.0351\times {10}^8\text{M}\left(1{\text{M=molL}}^{-1}\right)\\ \left[C_t\right]=3.51\times {10}^{-10}\text{M}\end{array}$

The concentration ofH₂O was found to be $3.51\times {10}^{-10}\text{M}$