A) 75
B) 50
C) 25
D) 100
Correct Answer: A
Solution :
\[{{N}_{2}}+3{{H}_{2}}\]⇌\[2N{{H}_{3}}\] Initial conc. 1 3 0 at equilibrium 1-0.81 3-2.43 1.62 0.19 0.57 No. of moles of \[{{N}_{2}}=\frac{28}{28}=1\]mole No. of moles of \[{{H}_{2}}=\frac{6}{2}=3\]mole No. of moles of \[N{{H}_{3}}=\frac{27.54}{17}=1.62\]mole \[{{K}_{c}}=\frac{{{[N{{H}_{3}}]}^{2}}}{[{{N}_{2}}]\,\,{{[{{H}_{2}}]}^{3}}}=\frac{{{[1.62]}^{2}}}{[0.19]\,\,{{[0.57]}^{3}}}\]=75You need to login to perform this action.
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