Question

For the reaction, $H_2+I_2\stackrel{k_1}{\underset{k_2}{\rightleftharpoons }}2HI$. The rate law expression is:

• A. $[-\frac{1}{2}\frac{d\left[HI\right]}{dt}]=k_1\left[H_2\right]\left[I_2\right]$
• B. $[-\frac{1}{2}\frac{d\left[HI\right]}{dt}]=\frac{k_1{\left[HI\right]}^2}{k_2\left[H_2\right]\left[I_2\right]}$
• C. $[-\frac{1}{2}\frac{d\left[HI\right]}{dt}]=k_1\left[H_2\right]\left[I_2\right]-k_2{\left[HI\right]}^2$
• D. $[-\frac{1}{2}\frac{d\left[HI\right]}{dt}]=k_1{k}_2\left[H_2\right]\left[I_2\right]$

Answer 1

The correct option is C $[-\frac{1}{2}\frac{d\left[HI\right]}{dt}]=k_1\left[H_2\right]\left[I_2\right]-k_2{\left[HI\right]}^2$

Use steady state approximation.

$k_1\left[H_2\right]\left[I_2\right]=$ formation of $HI$.

$k_2{\left[HI\right]}^2=$ destruction of $HI$.

$\therefore \frac{1}{2}\frac{d\left[HI\right]}{dt}=k_1\left[H_2\right]\left[I_2\right]-k_2{\left[HI\right]}^2$