For the reaction in equilibrium- N2O4g - 2NO2gThe concentrations of N2O4 and NO2 at equilibrium are 4-8-10-2 and 1-2-10-2 mol L-1 respectively- The value of Kc for this reaction is-

The correct option is A 3×10^ 3 mol L^ 1 #160; N_2O_4(g) ⇌ 2NO_2(g) K_c = [NO_2]^2/[N_2O_4] = [1.2×10^ 2]^2/4.8×10^ 2 = 3×10^ 3 #160; ...

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Byju's Answer

Standard XII

Chemistry

Rate of Reaction

For the react...

Question

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:

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

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

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

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

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Solution

The correct option is A $3 \times 10^{- 3} m o l L^{- 1}$
$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$

$K_{c} = \frac{{[N O_{2}]}^{2}}{[N_{2} O_{4}]}$

$= \frac{{[1.2 \times 10^{- 2}]}^{2}}{4.8 \times 10^{- 2}} = 3 \times 10^{- 3} m o l L^{- 1}$

Hence, option A is correct.

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Similar questions

Q. For the reaction equilibrium

N_{2} O_{4} ⇌ 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 i t r e^{- 1}

respectively. The value of

K_{c}

for the reaction is:

Q. 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:

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:

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:

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:

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For the reaction in equilibrium, N2O4(g)⇌2NO2(g)The concentrations of N2O4 and NO2 at equilibrium are 4.8×10−2 and 1.2×10−2molL−1 respectively. The value of Kc for this reaction is:

The correct option is A 3×10−3molL−1 N2O4(g)⇌2NO2(g)Kc=[NO2]2[N2O4]=[1.2×10−2]24.8×10−2=3×10−3molL−1Hence, option A is correct.

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:

The correct option is A $3 \times 10^{- 3} m o l L^{- 1}$
$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$

$K_{c} = \frac{{[N O_{2}]}^{2}}{[N_{2} O_{4}]}$

$= \frac{{[1.2 \times 10^{- 2}]}^{2}}{4.8 \times 10^{- 2}} = 3 \times 10^{- 3} m o l L^{- 1}$

Hence, option A is correct.