Q.56. The reaction was initiated with the amount 'a' of A(g). At equilibrium it is found that the a

Q: What is the correct answer for JEE Main Chemistry Q56?

The correct answer is B. Sol. $A(g) \rightleftharpoons B(g) + C(g)$ Initial: a 0 0 At equilibrium: a-x x x Total moles at equilibrium = a-x+x+x = a+x. Partial pressure of B, $P_B = \frac{x}{a+x} P$. Partial pressure of C, $P_C = \frac{x}{a+x} P$. Partial pressure of A, $P_A = \frac{a-x}{a+x} P$. $K_p = \frac{P_B \times P_C}{P_A} = \frac{(\frac{x}{a+x} P) \times (\frac{x}{a+x} P)}{(\frac{a-x}{a+x} P)} = \frac{x^2 P^2}{(a+x

Q: Q.56. The reaction was initiated with the amount 'a' of A(g). At equilibrium it is found that the amount of A(g) remaining is (a -x) at a total pressure of p. The equilibrium constant K of the…

The correct answer is B. Sol. $A(g) \rightleftharpoons B(g) + C(g)$ Initial: a 0 0 At equilibrium: a-x x x Total moles at equilibrium = a-x+x+x = a+x. Partial pressure of B, $P_B = \frac{x}{a+x} P$. Partial pressure of C, $P_C = \frac{x}{a+x} P$. Partial pressure of A, $P_A = \frac{a-x}{a+x} P$. $K_p = \frac{P_B \times P_C}{P_A} = \frac{(\frac{x}{a+x} P) \times (\frac{x}{a+x} P)}{(\frac{a-x}{a+x} P)} = \frac{x^2 P^2}{(a+x

Q: What are the options for this JEE Main Chemistry question?

A. $ \frac{x^2}{(a^2+x^2)xp}$ · B. $ \frac{x^2p}{a^2-x^2}$ · C. $ \frac{x^2}{a-x^2}2xp$ · D. $ \frac{x^2}{a-x}xp$

Question

Q.56. The reaction  was initiated with the amount 'a' of A(g). At equilibrium it is found that the amount of A(g) remaining is (a -x) at a total pressure of p. The equilibrium constant K of the reaction can be calculated from the expression :

FirstInTest · Accepted Answer

Sol. $A(g) ightleftharpoons B(g) + C(g)$ Initial: a 0 0 At equilibrium: a-x x x Total moles at equilibrium = a-x+x+x = a+x. Partial pressure of B, $P_B = \frac{x}{a+x} P$. Partial pressure of C, $P_C = \frac{x}{a+x} P$. Partial pressure of A, $P_A = \frac{a-x}{a+x} P$. $K_p = \frac{P_B 	imes P_C}{P_A} = \frac{(\frac{x}{a+x} P) 	imes (\frac{x}{a+x} P)}{(\frac{a-x}{a+x} P)} = \frac{x^2 P^2}{(a+x)^2} 	imes \frac{a+x}{P(a-x)} = \frac{x^2 P}{(a+x)(a-x)} = \frac{x^2 P}{a^2-x^2}$.

Answer

$ \frac{x^2}{(a^2+x^2)xp}$

Answer

$ \frac{x^2p}{a^2-x^2}$

Answer

$ \frac{x^2}{a-x^2}2xp$

Answer

$ \frac{x^2}{a-x}xp$

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