You visited us 2 times! Enjoying our articles? Unlock Full Access!

Law of Mass Action

The rate law ...

Question

The rate law for the decomposition of gaseous $N_{2} O_{5}$ , $N_{2} O_{5} \to 2 N O_{2} + \frac{1}{2} O_{2}$ , is observed to be:
$- \frac{d [N_{2} O_{5}]}{d t} = K [N_{2} O_{5}]$
A reaction mechanism which has been suggested to be consistent with this is:
$N_{2} O_{5} k ⇌ N O_{2} + N O_{3}$ (fast)
$N O_{2} + N O_{3} k_{1} - ---- \to N O_{2} + N O + O_{2}$ (slow)
$N O + N O_{3} k_{2} - ---- \to 2 N O_{2}$ (fast)
Show that the mechanism is consistent with the observed law.

Open in App

Solution

Given reaction :
$N_{2} O_{5} k ⇌ N O_{2} + N O_{3}$ (fast)...(i)
$N O_{2} + N O_{3}_{1} N O_{2} + N O + O_{2}$ (slow)...(ii)
$N O + N O_{3}_{2} 2 N O_{2}$ (fast)...(iii)
To Prove
$\frac{- d [N_{2} O_{5}]}{d t} = k [N_{2} O_{5}]$
Let us consider $k_{f}$ and $k_{b}$ are the rate constant
for forward and backward reaction
Assume $k = \frac{k_{f}}{K_{b}}$
Rate of the slow step $=$ rate of the reaction
$R = K_{1} [N O_{2}] [N O_{3}] . . . (i v)$
$[N O_{3}]$ is an intermediate Hence we apply the
steady state approximation
$\frac{d [N O_{3}]}{d t} = k_{f} [N_{2} O_{5}] + K_{b} [N O_{2}] [N O_{3}] - k_{1} [N O_{2}] [N O_{3}] - K_{2} [N O] [N O_{3}] = 0$
$k_{f} [N_{2} O_{5}] = k_{b} [N O_{2}] [N O_{3}] + k_{1} [N O_{2}] [N O_{3}] + K_{2} [N O] [N O_{3}]$
$[N O_{3}] = \frac{k_{f} [N_{2} O_{5}]}{k_{b} [N O_{2}] + k_{1} [N O_{2}] + k_{2} [N O_{2}]}$
Putting the value in eqn (iv)
$R = \frac{k_{1} [N O_{2}] k_{f} [N_{2} O_{5}]}{k_{b} [N O_{2} + k_{1} [N O_{2}] + k_{2} [N O]}$
Concentration of $[N O]$ and $[N O_{2}] k_{1}$ can be neglected because
they are generated in slow step
$\frac{d R}{d t} = \frac{k_{1} [N O_{2}] k_{f} [N_{2} O_{5}]}{k_{b} [N O_{2}]} = k_{1} [N_{2} O_{5}] \frac{k_{f}}{k_{b}} = k_{1} k [N_{2} O_{5}] = k [N_{2} O_{5}]$

Suggest Corrections

Similar questions

2 N_{2} O_{5} ⟶ 4 N O_{2} + O_{2}

N_{2} O_{5} ⇌ N O_{2} + N O_{3}

N O_{2} + N O_{3} ⟶ N O + N O_{2} + O_{2}

N O + N O_{3} ⟶ 2 N O_{2}

N_{2} O_{5} ⟶ 2 N O_{2} + 1 / 2 O_{2}

- \frac{d [N_{2} O_{5}]}{d t} = k_{1} [N_{2} O_{5}]

\frac{d [N O_{2}]}{d t} = k_{2} [N_{2} O_{5}]

\frac{d [O_{2}]}{d t} = k_{3} [N_{2} O_{5}]

k_{1}, k_{2}

k_{3}

N_{2} O_{5} \to 2 N O_{2} + \frac{1}{2} O_{2},

\frac{- d}{d t} [N_{2} O_{5}] = k_{1} [N_{2} O_{5}], \frac{d}{d t} [N O_{2}] = k_{2} [N_{2} O_{5}] a n d

\frac{d}{d t} [O_{2}] = k_{3} [N_{2} O_{3}],

N_{2} O_{5} \to 2 N O_{2} + \frac{1}{2} O_{2} .

G i v e n : - \frac{d [N_{2} O_{5}]}{d t} = K_{1} [N_{2} O_{5}]

\frac{d [N O_{2}]}{d t} = K_{2} [N_{2} O_{5}]

A n d \frac{d [O_{2}]}{d t} = K_{3} [N_{2} O_{5}]

The relation between $K_{1}, K_{2}$ and $K_{3}$ is

2 N_{2} O_{5} ⟶ 4 N O_{2} + O_{2}

\frac{- d [N_{2} O_{5}]}{d t} = k [N_{2} O_{5}]

\frac{d [N O_{2}]}{d t} = k^{'} [N_{2} O_{5}]

\frac{d [O_{2}]}{d t} = k^{''} [N_{2} O_{5}]

k

k^{'}

k

k^{''}

Join BYJU'S Learning Program