In a reaction A+2B ⇋ 2C; 2 moles ofA, 3 m

Discussion Forum : Chemical Equilibrium

Question -

In a reaction $A + 2 B ⇋ 2 C; 2 m o l e s o f A, 3$ mole of $B a n d 2 m o l e o f C$ are placed in a 2 L flask and the equilibrium concentration of $C i s 0.5 m o l e / L .$ The equilibrium constant K for the reaction is:

Options:

A . 0.073

B . 0.147

C . 0.05

D . 0.026

Answer: Option C
:
C

\begin{matrix} A + & 2 B ⇋ & 2 C I n i t i a l m o l e s & 2 & 3 & 2 M o l a r c o n c . & 2 / 2 = 1 m o l / L & 3 / 2 = 1.5 m o l / L & 2 / 2 = 1 m o l / L A t E q u i . C o n c & 1 + 0.25 = 1.25 m o l / L & 1.5 + 0.5 = 2 m o l / L & 0.5 m o l / L \end{matrix}

K = (0.5)^{2} / (1.25 \times (2)^{2}) = 0.05

Note that concentration of C has decreased, i.e., to attain equilibrium, reaction has shifted backwards. Decrease in conc. of C will be

0.5 m o l / L

and hence increase in concentration of B will be

0.5 m o l / L

and that of A will be

0.25 m o l / L

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