SATHEE: Equilibrium Question 599

Equilibrium Question 599

Question: $A_{(g)} + 3 B_{(g)}$ ⇌ $4 C_{(g)}$ . Starting concentration of A is equal to B, equilibrium concentration of A and C are same. $K_{c} =$ [Kerala CET 2005]

Options:

A) 0.08

B) 0.8

C) 8

D) 80

E) 1/8

Show Answer

Answer:

Correct Answer: C

Solution:

A + 3B ⇌ 4C a b 0 (a - x) (b - 3x) 4x $K_{C} = \frac{{[C]}^{4}}{[A] {[B]}^{3}} = \frac{4 x .4 x .4 x .4 x}{(a - x) (b - 3 x)}$ Given $a = b,, a - x = 4 x \Rightarrow a = 5 x = b$

$K_{C} = \frac{4 x .4 x .4 x .4 x}{(5 x - x) (5 x - 3 x)} = \frac{4 x .4 x .4 x .4 x}{4 x .2 x .2 x .2 x} = 8$ .

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Equilibrium Question 599

Question: A(g)+3B(g) ⇌ 4C(g) . Starting concentration of A is equal to B, equilibrium concentration of A and C are same. Kc= [Kerala CET 2005]

Options:

Answer:

Solution:

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Question: $A_{(g)} + 3 B_{(g)}$ ⇌ $4 C_{(g)}$ . Starting concentration of A is equal to B, equilibrium concentration of A and C are same. $K_{c} =$ [Kerala CET 2005]