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Derivation of Kp and Kc

For the react...

Question

For the reaction $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ , which of the following factors will have no effect on the value of equilibrium constant?

A

Temperature

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B

Initial concentration of

N_{2} O_{4}

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C

Pressure of catalyst

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D

Pressure

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Solution

The correct options are
A Initial concentration of $N_{2} O_{4}$
C Pressure of catalyst
D Pressure
Only a temprature change can force a equilibrium constant to change. If we can the pressure or the initial concentration the equilibrium concentration of reactant and product may change but equlibrium constant remains the same.

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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}

N O_{2}

at equilibrium are

4.8 \times 10^{- 2}

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 equilibrium:

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

, the concentration of

N_{2} O_{4}

N O_{2}

at equilibrium are

4.8 \times 10^{- 2}

1.2 \times 10^{- 2}

mol/L 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}

N O_{2}

at equilibrium are

4.8 \times 10^{- 2}

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

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}

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at equilibrium are

4.8 \times 10^{- 2}

1.2 \times 10^{- 2} mol/L

respectively. The value of

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for the reaction is:

Q. If, in the reaction

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g),

N_{2} O_{4}

taken initially and

x

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N_{2} O_{4}

which dissociates, then the number of moles at equilibrium will be:

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