A) 4
B) 27
C) 4/27
D) 1/27
Correct Answer: B
Solution :
Key Idea: Use law of mass action to find \[{{k}_{p}}\]for the reaction. \[N{{H}_{4}}COON{{H}_{2}}(s)2N{{H}_{3}}(g)+C{{O}_{2}}(g)\] \[\therefore \] \[{{K}_{P}}=\frac{p_{N{{H}_{3}}}^{2}\times {{p}_{C{{O}_{2}}}}}{{{p}_{N{{H}_{4}}COON{{H}_{2}}(s)}}}\] But \[{{p}_{N{{H}_{4}}COON{{H}_{2}}(s)}}=1\] (\[\because \] P of solids is taken as one) \[\therefore \] \[{{K}_{p}}={{p}^{2}}_{N{{H}_{3}}}\times {{p}_{C{{O}_{2}}}}\] \[={{(3)}^{2}}\times 3\] = 27You need to login to perform this action.
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