Find \( \frac{K_{P_{2}^{2}}}{K_{P_{1}^{6}}} \) if 0.25 moles of \( N_{2} \) are found finally.

Question

\( \rightarrow \) Class 11 CHEMISTRY \( NH _{-} \)(4) HS(s)hArrNh_(... \[ \begin{array}{l} N H_{4} H S(s) \Leftrightarrow N h_{3}(g)+H_{2} S(g) K_{P_{1}} \\ N H_{3}(g) \Leftrightarrow \frac{1}{2} N_{2}(g)+\frac{3}{2} H_{2}(g) K_{P_{2}} \end{array} \] \( 2 mol NH _{4} HS ( s ) \) is taken \& \( 50 \% \) of this is dissociated till at equilibrium in 1 litre container. Find \( \frac{K_{P_{2}^{2}}}{K_{P_{1}^{6}}} \) if 0.25 moles of \( N_{2} \) are found finally.

Answer 1

Correct answer: 27

\(NH_4HS(s) \rightleftharpoons \underset{1-0.5}{NH_3(g)} + \underset{1}{H_2S(g)}\)

\(\underset{0.5}{NH_3(g)} \rightleftharpoons \underset{0.25}{\frac 12N_2(g)} + \underset{0.75}{\frac 32 H_2(g)}\)

\(K_{P_1} = 0.5\)

\(K_{P_2} = \frac{(0.25^{1/2}) (0.75)^{3/2}}{0.5}\)