A) 0.128
B) 0.426
C) \[4.76\times {{10}^{-3}}\]
D) \[2.24\times {{10}^{-2}}\]
Correct Answer: C
Solution :
[c] \[\underset{Eq.\text{ }pressure}{\mathop{N{{H}_{4}}COON{{H}_{2}}(s)}}\,\underset{-2P}{\mathop{2N{{H}_{3}}(g)}}\,+\underset{P}{\mathop{C{{O}_{2}}(g)}}\,\] \[3P=0.318\Rightarrow P=0.106\] \[{{P}_{N{{H}_{3}}}}=2\times 0.106=0.212\] \[{{P}_{C{{O}_{3}}}}=0.106\] \[{{k}_{p}}=P_{N{{h}_{3}}}^{2}.{{P}_{C{{O}_{2}}}}=0.212\times 0.212\times 0.106\]You need to login to perform this action.
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