The following results have been obtained during the kinetic studies of the reaction: $2 \ NO + 2 \ H_2 \longrightarrow N_2 + 2 \ H_2O$
Expt$\frac{-d[NO]}{dt} \ (mol \ L^{-1} \ s^{-1})$$[NO] \ (mol \ L^{-1})$$[H_2] \ (mol \ L^{-1})$
$1$$4.8 \times 10^{-5}$$1 \times 10^{-2}$$1 \times 10^{-3}$
$2$$43.2 \times 10^{-5}$$3 \times 10^{-2}$$1 \times 10^{-3}$
$3$$86.4 \times 10^{-5}$$3 \times 10^{-2}$$2 \times 10^{-3}$

  • A
    $\frac{-d[NO]}{dt} = k[NO]^2[H_2]$
  • B
    $\frac{-d[NO]}{dt} = k[NO]^2[H_2]^{\frac{1}{2}}$
  • C
    $\frac{-d[NO]}{dt} = k[NO][H_2]^2$
  • D
    $\frac{-d[NO]}{dt} = k[NO][H_2]$

Explore More

Similar Questions

$A$ complex reaction takes place in the following steps:
$NO_2Cl_{(g)} \longrightarrow NO_{2(g)} + Cl_{(g)}$ (slow)
$NO_2Cl_{(g)} + Cl_{(g)} \longrightarrow NO_{2(g)} + Cl_{2(g)}$ (fast)
Identify the rate law equation for this reaction.

If a reaction has the experimental rate expression $\text{rate} = K [A]^2[B]$,what happens to the reaction rate if the concentration of $A$ is doubled and the concentration of $B$ is halved?

For the reaction $A + B \longrightarrow \text{product}$,the rate law equation is $\text{rate} = k[A]^2[B]$. If the rate of reaction is $0.22 \ mol \ L^{-1} \ s^{-1}$,calculate the rate constant $k$. Given: $[A] = 1 \ mol \ L^{-1}, [B] = 0.25 \ mol \ L^{-1}$.

What is the unit of the rate constant for a $4^{th}$ order reaction?

The numerical values of rate constants are same for first,second and third order reactions. Which one is true at a moment for rate of these three reactions if concentration of reactants is same and lesser than $1 \ M$?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo