A) 4
B) 8
C) 16
D) 32
Correct Answer: D
Solution :
[d] \[{{P}_{total}}={{P}_{HN{{O}_{3}}}}+{{P}_{N{{O}_{2}}}}+{{P}_{{{H}_{2}}O}}+{{P}_{{{O}_{2}}}}\] \[\therefore {{P}_{N{{O}_{2}}}}=4{{P}_{{{O}_{2}}}}\,and\,{{P}_{{{H}_{2}}O}}=2{{P}_{{{O}_{2}}}}\] \[\therefore \text{ }{{P}_{total}}={{P}_{HN{{O}_{3}}}}+7P{{o}_{2}}\] \[\Rightarrow 30-2=P{{o}_{2}}\times 7\] \[{{K}_{p}}=\frac{P_{N{{O}_{2}}}^{4}.{{P}_{{{H}_{2}}O}}.P{{O}_{2}}}{P_{HN{{O}_{3}}}^{4}}\] \[={{\frac{{{(4\times 4)}^{4}}\times {{(2\times 4)}^{2}}\times 4}{{{2}^{4}}}}^{3}}={{2}^{20}}\] \[{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta {{n}_{g}}}}={{K}_{c}}{{(0.08\times 400)}^{3}}\] \[\Rightarrow {{K}_{c}}=\frac{{{2}^{20}}}{{{(32)}^{3}}}=32\]You need to login to perform this action.
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