For the reaction at equilbrium- N2O4 g- 2NO2g- the concentration of N2O4 and NO2 at equilibrium are 4-8- 10-2 and 1-2 - 10-2mol-L respectively- The value of Kc for the reaction is-

The equilibrium reaction is N_2O_4 (g)⇌ 2NO_2(g) K_c = [NO_2]^2[N_2O_4] = (1.2×10^ 2)^24.8×10^ 2 ⇒ K_c = 3×10^ 3M ...

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Standard XII

Chemistry

Equilibrium Constant and Standard Free Energy Change

For the react...

Question

For the reaction at equilbrium: $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ , the concentration of $N_{2} O_{4}$ and $N O_{2}$ at equilibrium are $4.8 \times 10^{- 2}$ and $1.2 \times 10^{- 2} mol/L$ respectively. The value of $K_{c}$ for the reaction is:

3 \times 10^{- 3} M

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3 \times 10^{3} M

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3.3 \times 10^{2} M

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3 \times 10^{- 1} M

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Solution

The correct option is A $3 \times 10^{- 3} M$
The equilibrium reaction is $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$
$K_{c} = \frac{[N O_{2}]^{2}}{[N_{2} O_{4}]} = \frac{(1.2 \times 10^{- 2})^{2}}{4.8 \times 10^{- 2}}$
$\Rightarrow K_{c} = 3 \times 10^{- 3} M$

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Similar questions

Q. For a reaction equilibrium,

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

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4.8 \times 10^{- 2}

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the concentrations of

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Q. The density of an equilibrium mixture of

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For the reaction at equilbrium: N2O4(g)⇌2NO2(g), the concentration of N2O4 and NO2 at equilibrium are 4.8×10−2 and 1.2×10−2mol/L respectively. The value of Kc for the reaction is:

The correct option is A 3×10−3MThe equilibrium reaction is N2O4(g)⇌2NO2(g) Kc=[NO2]2[N2O4]=(1.2×10−2)24.8×10−2 ⇒Kc=3×10−3M

For the reaction at equilbrium: $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ , the concentration of $N_{2} O_{4}$ and $N O_{2}$ at equilibrium are $4.8 \times 10^{- 2}$ and $1.2 \times 10^{- 2} mol/L$ respectively. The value of $K_{c}$ for the reaction is:

The correct option is A $3 \times 10^{- 3} M$
The equilibrium reaction is $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$
$K_{c} = \frac{[N O_{2}]^{2}}{[N_{2} O_{4}]} = \frac{(1.2 \times 10^{- 2})^{2}}{4.8 \times 10^{- 2}}$
$\Rightarrow K_{c} = 3 \times 10^{- 3} M$