For a reaction equilibrium- N2 O4g- 2NO2 g- the concentrations of N2O4 and NO2 at equilibrium are 4-8- 10-2 and 1-2- 10-2 mol-L respectively- The value of Kc- for the reaction is-

The correct option is A 3× 10^ 3 mol/L According law #10;of active mass, #10; K_c =[NO_2]^2[N_2O_4] = [1.2× 10^ 2]^24.8× #10;10^ 2 #10; ...

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Byju's Answer

Standard XII

Chemistry

Characteristics of Equilibrium Constant

For a reactio...

Question

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

3 \times 10^{- 3} m o l / L

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3.3 \times 10^{- 3} m o l / L

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3 \times 10^{- 1} m o l / L

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3.3 \times 10^{- 1} m o l / L

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Solution

The correct option is A $3 \times 10^{- 3} m o l / L$
According law of active mass,
$K_{c} = \frac{[N O_{2}]^{2}}{[N_{2} O_{4}]}$ $= \frac{[1.2 \times 10^{- 2}]^{2}}{4.8 \times 10^{- 2}}$

$= 0.3 \times 10^{- 2} = 3 \times 10^{- 3}$

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Similar questions

Q. For the reaction equilibrium

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

the concentrations of

N_{2} O_{4}

and

N O_{2}

at equilibrium are

4.8 \times 10^{- 2}

and

1.2 \times 10^{- 2} m o l L^{- 1}

respectively. The value of

K_{c}

for the reaction is

Q. 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:

Q. For the reaction at equilibrium:

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:

Q. For the reaction in equilibrium,

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

The concentrations of

N_{2} O_{4}

and

N O_{2}

at equilibrium are

4.8 \times 10^{- 2}

and

1.2 \times 10^{- 2} m o l L^{- 1}

respectively. The value of

K_{c}

for this reaction is:

P C l_{5}, P C l_{3} a n d C l_{2}

are at equilibrium at

500

K in a closed container and their concentrations are

0.8 \times 10^{- 3} m o l L^{- 1}, 1.2 \times 10^{- 3} m o l L^{- 1}

and

1.2 \times 10^{- 3} m o l L^{- 1}

respectively.
The value of

K_{c}

for the reaction:

P C l_{5 (g)} ⇌ P C l_{3 (g)} + C l_{2 (g)}

will be:

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

The correct option is A 3×10−3 mol/LAccording law of active mass, Kc=[NO2]2[N2O4]=[1.2×10−2]24.8×10−2 =0.3×10−2=3×10−3

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

The correct option is A $3 \times 10^{- 3} m o l / L$
According law of active mass,
$K_{c} = \frac{[N O_{2}]^{2}}{[N_{2} O_{4}]}$ $= \frac{[1.2 \times 10^{- 2}]^{2}}{4.8 \times 10^{- 2}}$

$= 0.3 \times 10^{- 2} = 3 \times 10^{- 3}$