Question

For the reaction,
$2NO\left(g\right)+2H_2\left(g\right)\to N_2\left(g\right)+2H_{2}O\left(g\right)$
The rate expression can be written in the following ways
$\frac{d\left[N_2\right]}{dt}=k_1\left[NO\right]\left[H_2\right];-\frac{d\left[H_2\right]}{dt}=k_2\left[NO\right]\left[H_2\right]$
$-\frac{d\left[NO\right]}{dt}=k_3\left[NO\right]\left[H_2\right];-\frac{d\left[H_2\right]}{dt}=k_4\left[NO\right]\left[H_2\right]$
The relationship between $k_1,k_2,k_2,k_4,$ is

• A. $k_2=k_1=k_3=k_4$
• B. $k_2=2k_1=k_3=k_4$
• C. $k_2=2k_3=k_1=k_4$
• D. $k_2=k_1=k_3=2k_4$

Answer 1

Answer: (B)

$2NO+2H_2\to N_2+2H_{2}O$
$\frac{r_{NO}}{2}=\frac{r_{H_2}}{2}{=}^r\frac{r_{N_2}}{1}=\frac{r_{H_{2}O}}{2}$
$\frac{k_3\times \left[NO\right]\times \left[H_2\right]}{2}=\frac{k_4\times \left[NO\right]\times \left[H_2\right]}{1}$
$=\frac{k_2\times \left[NO\right]\times \left[H_2\right]}{2}$
$\therefore k_3=k_4=2k_1=k_2$