For an elementary gaseous phase reaction :2NO+O2⇌2NO2 at 27∘CRate of the forward reaction is given by rate =2 × 103NO2O2and rate of reverse reaction at is given by rate =20NO22Hence, equilibrium constant for reactionNO2⇌NO+12 at 27∘Cwill be:

rf=2×103NO2O2rb=20NO22At equilibrium rf=rb⇒K=NO22NO2O2=2×10320=100K′=1K=110=0.1So, equilibrium constant for reaction is 0.1

For an elementary gaseous phase reaction :2NO+O2⇌2NO2 at 27∘CRate of the forward reaction is given by rate =2 × 103NO2O2and rate of reverse reaction at is given by rate =20NO22Hence, equilibrium constant for reactionNO2⇌NO+12 at 27∘Cwill be:

rf=2×103NO2O2rb=20NO22At equilibrium rf=rb⇒K=NO22NO2O2=2×10320=100K′=1K=110=0.1So, equilibrium constant for reaction is 0.1

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For an elementary gaseous phase reaction :
$2 NO + O_{2} ⇌ 2 {NO}_{2}$ at $27^{\circ} C$
Rate of the forward reaction is given by rate $= 2 \times 10^{3} [{NO}^{2} [O_{2}]$ and rate of reverse reaction at is given by rate $= 20 {[{NO}_{2}]}^{2}$
Hence, equilibrium constant for reaction
${NO}_{2} ⇌ NO + \frac{1}{2}$ at $27^{\circ} C$
will be:

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100

0.1

0.01

answer is C.

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Detailed Solution

$\begin{array}{l} r_{f} = 2 \times 10^{3} [{NO}^{2} [O_{2}] \\ r_{b} = 20 {[{NO}_{2}]}^{2} \end{array}$

At equilibrium

$\begin{array}{l} r_{f} = r_{b} \Rightarrow K = \frac{{[{NO}_{2}]}^{2}}{[{NO}^{2} [O_{2}]} = \frac{2 \times 10^{3}}{20} = 100 \\ K^{'} = \frac{1}{\sqrt{K}} = \frac{1}{10} = 0.1 \end{array}$

So, equilibrium constant for reaction is 0.1

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For an elementary gaseous phase reaction :2NO+O2⇌2NO2 at 27∘CRate of the forward reaction is given by rate =2 × 103NO2O2and rate of reverse reaction at is given by rate =20NO22Hence, equilibrium constant for reactionNO2⇌NO+12 at 27∘Cwill be:

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