Equilibrium Question 771
Question: $ 28g,N_2 $ and 6.0 g of $ H_2 $ are heated over catalyst in a closed one litre flask of $ 450{}^\circ C $ . The entire equilibrium mixture required 500 mL of 1.0 M $ H_2SO_4 $ for neutralisation. The value of $ K_{c} $ for the reaction $ N_2(g)+3H_2(g)\rightarrow 2NH_3(g) $ is
Options:
A) $ 0.06mo{l^{-2}}L^{2} $
B) $ 0.59mo{l^{-2}}L^{2} $
C) $ 1.69mol^{2}{L^{-2}} $
D) $ 0.03mol^{2}{L^{-2}} $
Show Answer
Answer:
Correct Answer: B
Solution:
Moles of $ N_2=\frac{28}{28}=1, $ Moles of $ H_2=\frac{6}{2}=3 $ Moles of $ H_2SO_4 $ required $ =\frac{500\times 1}{1000}=0.5 $ Moles of $ NH_3 $ neutralised by $ H_2SO_4=1.0 $
$ ( 2NH_3+H_2SO_4\xrightarrow{{}}{{( NH_4 )}_2}SO_4 ) $ Hence 1 mol of $ NH_3 $ by the reaction between $ N_2 $ and $ H_2 $ . $ \begin{aligned} & \begin{matrix} {} & {} & {} & N_2+3H_2\rightarrow 2NH_3 \\ \end{matrix} \\ & ,\begin{matrix} initial & {} & 1 & {} \\ \end{matrix}\begin{matrix} 3 & {} & & 0 \\ \end{matrix} \\ & \begin{matrix} at,eqm & 1-0.5 & 3-0.5\times 3 & 1 \\ \end{matrix} \\ \end{aligned} $
$ K_{c}=\frac{1\times 1}{0.5\times {{(1.5)}^{3}}}=0.592 $
