Question

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

${\text{HO}}_2\left(g\right)+{\text{HO}}_2\left(g\right){\text{→H}}_{\text{2}}{\text{O}}_{\text{2}}\left(g\right)$

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

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

$$\begin{gathered} 2A\to \text{Products} \\ \text{Rate}=k{\left[A\right]}^2........\left(1\right) \end{gathered}$$

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

Rate expression:

The rate law for second order reaction is given as

$$\begin{gathered} \frac{1}{\left[C_0\right]}-\frac{1}{\left[C_t\right]}=kt............\left(2\right) \\ Where\left[C_0\right]=\text{initialconcendrationofreactents} \\ \left[C_t\right]\hspace{0.17em}=\text{Finalconcendrationofreactents} \\ K=ratecons\tan t \\ t=timeinsaeconds \\ Fromthegivendatasubstitutingthevaluesgiveninthequestioninequation-1 \\ \left[C_0\right]=2.0\times {10}^{-8}M \\ \left[C_t\right]=? \\ K=1.4\times {10}^{9}Lmo{l}^{-1}{s}^{-1} \\ t=\hspace{0.17em}10seaconds \\ \frac{1}{\left[{\text{C}}_t\right]}=\frac{1}{2.0\times {10}^{-8}}+2\times \left(1.4\times {10}^9{\text{molL}}^{-1}{\text{s}}^{\text{-1}}\right)\times \left(10s\right) \\ \frac{1}{\left[{\text{C}}_t\right]}=\left(\frac{1}{2.0\times {10}^{-8}}+2.0\times {10}^{-8}\right){\text{molL}}^{-1} \\ \frac{1}{\left[{\text{C}}_t\right]}\text{=}\frac{57}{2.0\times {10}^{-8}}{\text{molL}}^{-1} \\ \frac{1}{\left[{\text{C}}_t\right]}=28.5\times {10}^8{\text{molL}}^{-1} \\ \frac{1}{\left[{\text{C}}_t\right]}=\frac{1}{28.5\times {10}^8}{\text{molL}}^{-1} \\ \left[{\text{C}}_t\right]\hspace{0.17em}=0.0351\times {10}^8\text{M}\left(1\text{M}={\text{molL}}^{-1}\right) \\ \left[{\text{C}}_t\right]=3.51\times {10}^{-10}\text{M} \end{gathered}$$

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