1. Tardigrade
  2. Question
  3. Chemistry
  4. For the reaction, 2NO(g)+2H2(g)→ N2(g)+2H2O(g) The rate expression can be written in the following ways (d[N2]/dt)=k1[NO][H2];-(d[H2]/dt)=k2[NO][H2] -(d[NO]/dt)=k3[NO][H2];-(d[H2]/dt)=k4[NO][H2] The relationship between k1,k2,k2,k4, is

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Q. 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

Chemical Kinetics

Solution:

$2NO+2H_{2}\to N_{2}+2H_{2}O$
$\frac{r_{NO}}{2}=\frac{r_{H_{2}}}{2}=^{r}\frac{r_{N_{2}}}{1}= \frac{r_{H_{2}O}}{2}$
$\frac{k_{3}\times\left[NO\right]\times\left[H_{2}\right]}{2}=\frac{k_{4}\times\left[NO\right]\times\left[H_{2}\right]}{1}$
$=\frac{k_{2}\times\left[NO\right]\times\left[H_{2}\right]}{2}$
$\therefore k_{3}=k_{4}=2k_{1}=k_{2}$