Question
One mole each of He and $A(g)$ are taken in a 10 L closed flask and heated to 400 K to establish the following equilibrium.
$$A(g) \rightleftharpoons B(g) $$$K_{C}$ for this reaction at 400 K is 4.0. The partial pressures (in atm ) of He and $B(g)$ are respectively (at equilibrium).
(Assume He, $A(g)$ and $B(g)$ behave as ideal gases)
(Given : $R = 0.082\text{ }L\text{ }atm{\text{ }K}^{- 1}{\text{ }mol}^{- 1}$ )
Options
Solution
Sol. [Key Concept: For $A \rightleftharpoons \ B$, $K_{c} = \frac{x/10}{(1 - x)/10} = 4$ gives $x = 0.8$; then each partial pressure is $\frac{nRT}{V}$.]
$$\begin{matrix} & A \rightleftharpoons \text{ }B \\ & t = 0\ 1\text{ }mol - \\ & t_{{eq}^{m}}\ 1 - x\ x \\ & {\text{ }K}_{C} = \frac{x/10}{\frac{1 - x}{10}} = 4 \\ & x = 0.8 \\ & P_{He} = \frac{1 \times 0.082 \times 400}{10} = 3.28\text{ }atm \\ & P_{B} = \frac{0.8 \times 0.082 \times 400}{10} = 2.624\text{ }atm \end{matrix}$$
Create a free account to view solution
View Solution Free