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Derivation of Kp and Kc

26.In a rever...

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26.In a reversible reaction A(g)+B(g)--->C(g). Initially 1 mol of each reactant is taken in 2L container if at equilibrium 20% of A reacts with B. Then what will be equilibrium constant Kc for the reaction

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Q. The equilibrium constant

K_{p}

for the reaction

A (g) ⇌ B (g) + C (g)

27 C a n d 4 a t 47 .

For the reaction calculate enthalpy change for the

B (g) + C (g) ⇌ A (g)

(G i v e n : R = 2 c a l / m o l - K)

K_{C}

for the reaction,

A_{(g)} + B_{(g)} ⇌ C_{(g)} + D_{(g)}

20

25^{\circ} C

. If a container contains

1, 2, 4, 5

moles per litre of

A_{(g)}, B_{(g)}, C_{(g)}

D_{(g)}

respectively at

25^{\circ} C

, then the reaction shall:

The homogeneous reaction is carried out in a 2-litre container at a particular temperature by taking 1 mole each of A, B, C and D respectively. If

K_{C}

for the reaction is

\frac{1}{4}

then the equilibrium concentration of C is :

A (g) + B (g) ⇌ C (g) + D (g)

Q. In the reversible reaction

A + B ⇌ C + D

, the concentration of each

C

D

at equilibrium was

0.8

mole/litre, then the equilibrium constant

K_{c}

will be (if initial concentration of

A

B

1 M

Q. In the reversible reaction

2 N H_{3} (g) \Leftrightarrow N_{2} (g) + 3 H_{2} (g)

N H_{3}

is taken initially in a litre flask and 0.2 moles of

N_{2}

are formed at equilibrium. Its equilibrium constant

K_{c}

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Derivation of Kp and Kc

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