Question

The reaction of $O_3$ with chlorine atom is given as:
$O_3\left(g\right)+Cl\left(g\right)⟶O_2\left(g\right)+ClO\left(g\right);$ $k_1=5.2\times {10}^{9}L{mol}^{-1}{sec}^{-1}$
$ClO\left(g\right)+O\left(g\right)⟶Cl\left(g\right)+O_2\left(g\right);$ $k_2=2.6\times {10}^{10}L{mol}^{-1}{sec}^{-1}$
Which of these values is closest to the rate constant of the overall reaction?
$O_3\left(g\right)+O\left(g\right)⟶2O_2\left(g\right)$

• A. $5.2\times {10}^9$
• B. $2.6\times {10}^{10}$
• C. $3.1\times {10}^{10}$
• D. $1.4\times {10}^{20}$

Answer 1

Answer: (A)

$5.2\times {10}^9$

$k_1\left[O_3\right]\left[Cl\right]=\frac{d}{dt}\left[ClO\right]k_2\left[ClO\right]\left[O\right]=\frac{-d}{dt}\left[ClO\right]k_1\left[O_3\right]\left[Cl\right]=k_2\left[ClO\right]\left[O\right]\frac{k_1}{k_2}\frac{\left[O_3\right]\left[Cl\right]}{\left[O\right]}=\left[ClO\right]$

Similarly,

$\frac{k_2}{k_1}\frac{\left[ClO\right]\left[O\right]}{\left[O_3\right]}=\left[Cl\right]$

$\therefore$ We find that on combining

$O_3+Cl+ClO⟶O_2+ClO+Cl+O_2{O}_3+O⟶{2O}_2$

The net rate depends on consumption of $O_3$ only because O is an intermediate in the steady state assumption.

$\therefore k=k_1=5.2\times {10}^9$