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Derivation of Kp and Kc

K C for the r...

Question

$K_{c}$ for the reaction $N_{2} O_{4} ⇌ 2 N O_{2}$ in chloroform at 291 K is 1.14. Calculate the $K_{p}$ of the reaction when the concentration of the two gases are $0.5 m o l d m^{- 3}$ each at the same temperature. $(R = 0.082 L a t m K^{- 1} m o l^{- 1})$

A

-27.1

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B

-19.6

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C

1.1

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D

27.1

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Solution

The correct option is D 27.1
From the given data
$T = 291 K; R = 0.082 l i t a t m K^{- 1} m o l^{- 1}$
$K_{c} = 1.14; c = 0.5 m o l d m^{- 3}$
$△ n = 2 - 1 = 1$ in this case
we know the relation,
$K_{p} = K_{c} (R T)^{△ n}$
$K_{p} = 1.14 (0.082 \times 291) = 27.1$

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Similar questions

K_{c}

for the reaction

N_{2} O_{4} ⇌ 2 N O_{2}

in chloroform at 291 K is 1.14. Calculate the

K_{p}

of the reaction when the concentration of the two gases are

0.5 m o l d m^{- 3}

each at the same temperature.

(R = 0.082 L a t m K^{- 1} m o l^{- 1})

Q. Standard Free Energy and Equilibrium constant: The change in free energy for a reaction taking place between gaseous reactants and products represented by general equation.

Δ G = Δ G^{\circ} + R T l n Q_{P}

the condition for a system to be at equilibrium is that

Δ G = 0 and Q_{p} = K_{P}

Thus at equilibrium

Δ G^{\circ} = - R T l n K_{P}

Note: In the reaction, where all gaseous reactants and products;

K represents K_{P}

, we may conclude that for standard reactions, i.e., at

1 M or 1 atm

Δ G^{\circ} = - v e or K > 1 : forward reaction is feasible

Δ G^{\circ} = + v e or K < 1 : reverse reaction is feasible

Δ G^{\circ} = 0 or K = 1 : reaction is at equilibrium (very rare)

K_{c}

N_{2} O_{4} ⇌ 2 N O_{2}

in chloroform at

291 K

1.14

. Calculate the free energy change of the reaction when the concentration of the two gases are

0.5 {mol dm}^{- 3}

each at the same temperature. (

R = 0.082 {lit atm K}^{- 1} {mol}^{- 1}

)

k_{c}

for the reaction

N_{2} O_{4} (g) ⇋ 2 N O_{2}

4.63 \times 10^{- 3}

25^{o} C

a. what is the value of

K_{p}

at this temperature.
b. At

25^{o} C

, if the partial pressure of

N_{2} O_{4}

(g) at equilibrium is 0.2 atm, calculate equilibrium pressure of

N O_{2} (g)

Q. For the reaction

N_{2} O_{4} ⇌ 2 N O_{2}

, equilibrium mixture contains

N O_{2}

P = 1.1 a t m

N_{2} O_{4}

P = 0.28 a t m

at 350 K. The volume of the container is doubled. Calculate the equilibrium pressures of the two gases when the system reaches new equilibrium :

Q. For the reaction,

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g), K_{p} = 0.157 a t m

300 K

. Calculate the value of

K_{c}

for the same reaction at the same temperature.

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Derivation of Kp and Kc

Standard XII Chemistry

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