A) 4
B) 27
C) 4/27
D) 1/27
Correct Answer: B
Solution :
Equilibrium pressure = 3atm \[N{{H}_{4}}COON{{H}_{2(s)}}\] ⇌ \[2N{{H}_{3(g)}}+C{{O}_{2(g)}}\] \[{{K}_{p}}=p_{N{{H}_{3}}}^{2}.{{p}_{C{{O}_{2}}}}={{3}^{2}}.3=27\]You need to login to perform this action.
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