In the given reaction : A+2 B leftharpoons 2 C 2 moles each of A and B present in 10 ~L of solution combine to form 1 mole of C . Calculate K_c for the reaction.

Correct answer: B. 6.67 A+2 B leftharpoons 2 C As 1 mole of C is formed at equilibrium, 2 x=1 x=0.5 Number of moles of A at equilibrium =1.5, B=1 and C=1 Equilibrium concentration of A=1.5 / 10 gathered B=1 / 10 and C=1 / 10 K_c=([C]^2)/([A][B]^2)=([1 / 10]^2)/([1.5 / 10][1 / 10]^2)=6.67 gathered

AIIMS 201827 May

Exam: AIIMS
Year: 2018
Date: 27 May
Subject: Chemistry
Chapter: Chemical Equilibrium

AIIMS 2018 · 27 May · Chemistry · Chemical Equilibrium

In the given reaction : A+2 B leftharpoons 2 C 2 moles each of A...

In the given reaction :

A+2 B \rightleftharpoons 2 C

2 moles each of

A

and

B

present in

10 \mathrm{~L}

of solution combine to form 1 mole of

C

. Calculate

K_c

for the reaction.

Options

A $1.5$
B $6.67$
C $0.15$
D $2.3$

Correct answer

B. $6.67$

Step-by-step solution

A+2 B \rightleftharpoons 2 C

As 1 mole of

is formed at equilibrium, 2 x=1 \Rightarrow x=0.5 \Rightarrow

Number of moles of A

at equilibrium

=1.5, B=1

and

C=1 \therefore

Equilibrium concentration of A=1.5 / 10 \begin{gathered}B=1 / 10 \text{ and } C=1 / 10 \\ K_c=\frac{[C]^2}{[A][B]^2}=\frac{[1 / 10]^2}{[1.5 / 10][1 / 10]^2}=6.67\end{gathered}

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