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Characteristics of Equilibrium Constant

Consider the ...

Question

Consider the reactions,
${N O}_{2} (g) ⇌ \frac{1}{2} N_{2} (g) + O_{2} (g)$ ; $K_{1}$
$N_{2} O_{4} ⇌ 2 {N O}_{2}$ ; $K_{2}$
Give the equilibrium constant for the formation of $N_{2} O_{4}$ from $N_{2}$ and $O_{2}$ .

\frac{1}{K_{1}^{2}} \times \frac{1}{K_{2}}

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\frac{1}{2 K_{1}} + \frac{1}{K_{2}}

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\sqrt{\frac{1}{{K_{1} K}_{2}}}

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\frac{K_{2}}{K_{1}}

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Solution

The correct option is A $\frac{1}{K_{1}^{2}} \times \frac{1}{K_{2}}$
Consider the reactions
${N O}_{2} (g) ⇌ \frac{1}{2} N_{2} (g) + O_{2} (g)$ $K_{1}$ ....(1)
$N_{2} O_{4} ⇌ 2 {N O}_{2}$ $K_{2}$ .....(2)
Multiply equation (1) with 2
$2 {N O}_{2} (g) ⇌ N_{2} (g) + 2 O_{2} (g)$ $K_{1}$ $K_{3} = K_{1}^{2}$
Reverse the reaction
$N_{2} (g) + 2 O_{2} (g)$ $K_{1} ⇌ 2 {N O}_{2} (g)$ $K_{4} = (\frac{1}{K_{1}})^{2}$ .....(3)
Reverse the equation (2)
$2 {N O}_{2} ⇌ N_{2} O_{4}$ $K_{5} = \frac{1}{K_{2}}$ .....(4)
Multiply the equations (3) and (4)
The equilibrium constant for the formation of $N_{2} O_{4}$ from $N_{2}$ and $O_{2}$ is $K_{4} \times K_{5} = \frac{1}{K_{1}^{2}} \times \frac{1}{K_{2}}$

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Similar questions

N O_{2} ⇌ \frac{1}{2} N_{2} + O_{2}, K_{1}

N_{2} O_{4} ⇌ 2 N O_{2}, K_{2}

N_{2} O_{4}

N_{2}

O_{2}

K

K_{1}

K_{2}

N_{2} + 2 O_{2} ⇋ 2 N O_{2}, K

N_{2} + O_{2} ⇋ 2 N O, K_{1}

2 N O + O_{2} ⇋ 2 N O_{2}, K_{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} .

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

N_{2} O_{5} ⟶ 2 {N O}_{2} + \frac{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 O}_{2}]

- \frac{d [O_{2}]}{d t} = K_{3} [O_{2}]

K_{1}, K_{2}, K_{3}

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