Question
In Haber process $$30L$$ of dihydrogen and $$30L$$ of dinitrogen were taken for reaction which yielded only $$50\% $$ of the expected product. What will be the composition of gaseous mixture under the aforesaid condition in the end?
A.
$$20$$ $$L$$ ammonia, $$10$$ $$L$$ nitrogen, $$30$$ $$L$$ hydrogen
B.
$$20$$ $$L$$ ammonia, $$25$$ $$L$$ nitrogen, $$15$$ $$L$$ hydrogen
C.
$$20$$ $$L$$ ammonia, $$20$$ $$L$$ nitrogen, $$20$$ $$L$$ hydrogen
D.
$$10$$ $$L$$ ammonia, $$25$$ $$L$$ nitrogen, $$15$$ $$L$$ hydrogen
Answer :
$$10$$ $$L$$ ammonia, $$25$$ $$L$$ nitrogen, $$15$$ $$L$$ hydrogen
Solution :
\[\underset{\begin{smallmatrix}
1\,V \\
10\,L
\end{smallmatrix}}{\mathop{{{N}_{2}}}}\,+\underset{\begin{smallmatrix}
3\,V \\
30\,L
\end{smallmatrix}}{\mathop{3{{H}_{2}}}}\,\to \underset{\begin{smallmatrix}
2\,V \\
20\,L
\end{smallmatrix}}{\mathop{2N{{H}_{3}}}}\,\]
As only $$50\% $$ of the expected product is formed, hence only $$10 L$$ of \[N{{H}_{3}}\] is formed.
Thus, for the production of $$10 L$$ of \[N{{H}_{3}}\] , $$5L$$ of $${N_2}$$ and $$15L$$ of $${H_2}$$ are used and composition of gaseous
mixture under the aforesaid condition in the end is
$$\eqalign{
& {H_2} = 30 - 15 = 15\,L \cr
& {N_2} = 30 - 5 = 25\,L \cr
& N{H_3} = 10L \cr} $$