For the reaction 2 H _ 2 (g) + 2NO(g) → N _ 2 (g) + 2 H _ 2 O(g) the observed rate expression is, rate = k _ f [ NO ] ^ 2 [ H _ 2 ] . The rate expression for the reverse reaction is:

Correct answer: D. k _ b [ N _ 2 ] [ H _ 2 O] ^ 2 / [ H _ 2 ] K _ eq = k _ f k _ b = [ N _ 2 ] [ H _ 2 O] ^ 2 [ H _ 2 ] ^ 2 [ NO ] ^ 2 At equilibrium r _ f = r _ b k _ f [ H _ 2 ] [ NO ] ^ 2 = k _ b [ N _ 2 ] [ H _ 2 O] ^ 2 [ H _ 2 ] Hence, rate expression for reverse reaction. = k _ b [ N _ 2 ] [ H _ 2 O] ^ 2 [ H _ 2 ]

JEE Main 20207 JanuaryEvening Shift

Exam: JEE Main
Year: 2020
Date: 7 January
Shift: Evening Shift
Subject: Chemistry
Chapter: Chemical Kinetics

JEE Main 2020 · 7 January · Evening Shift · Chemistry · Chemical Kinetics

For the reaction 2 H _ 2 (g) + 2NO(g) → N _ 2 (g) + 2 H _ 2 O(g)...

For the reaction

2{H}_{2}\left(g\right) + 2NO\left(g\right) \to {N}_{2}\left(g\right) + 2{H}_{2}O\left(g\right)

the observed rate expression is, rate

= {k}_{f}{\left[\text{ NO }\right]}

^{2}

\left[{H}_{2}\right]

. The rate expression for the reverse reaction is:

Options

A ${\text{k}}_{\text{b}}$ \left[{\text{N}}_{2}\right]{ $\left[{\text{H}}_{2}\text{O}\right]}^{2}$
B ${k}_{b}\left[{N}_{2}\right]{\left[{H}_{2}O\right]}^{2} / \left[\text{NO}\right]$
C ${\text{k}}_{\text{b}}\left[{\text{N}}_{2}\right]\left[{\text{H}}_{2}\text{O}\right]$
D ${k}_{b}\left[{N}_{2}\right]{\left[{H}_{2}O\right]}^{2} / \left[{\text{H}}_{2}\right]$

Correct answer

D. ${k}_{b}\left[{N}_{2}\right]{\left[{H}_{2}O\right]}^{2} / \left[{\text{H}}_{2}\right]$

Step-by-step solution

{\text{K}}_{\text{eq}} = \dfrac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{b}}} = \dfrac{\left[{\text{N}}_{2}\right]{\left[{\text{H}}_{2} \text{O}\right]}^{2}}{{\left[{\text{H}}_{2}\right]}^{2} {[ \text{NO} ]}^{2}}

At equilibrium

{\text{r}}_{\text{f}} = {\text{r}}_{\text{b}}

{\text{k}}_{\text{f}} \left[{\text{H}}_{2}\right]{[ \text{NO} ]}^{2} = {\text{k}}_{\text{b}} \dfrac{\left[{\text{N}}_{2}\right]{\left[{\text{H}}_{2} \text{O}\right]}^{2}}{\left[{\text{H}}_{2}\right]}

<br /><br />Hence, rate expression for reverse reaction.<br /><br />

= {\text{k}}_{\text{b}} \dfrac{

\left[{\text{N}}_{2}\right]{\left[{\text{H}}_{2} \text{O}\right]}^{2}}{\left[{\text{H}}_{2}\right]}

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