| The values of \[{{K}_{\text{P}}}\text{/}{{K}_{c}}\] for the following reactions at 300 K are respectively: |
| \[(At\,300\,K,\,Rt=24.62\,d{{m}^{3}}atm\,\,mo{{l}^{-1}})\] |
| \[{{N}_{2}}(g)+{{O}_{2}}(g)=2NO(g)\] |
| \[{{N}_{2}}{{O}_{4}}(g)=2N{{O}_{2}}(g)\] |
| \[{{N}_{2}}(g)+3{{H}_{2}}(g)=2N{{H}_{3}}(g)\] |
A) 1, 24.62 \[d{{m}^{3}}\] atm \[mo{{l}^{-1}},606.0d{{m}^{6}}\] \[at{{m}^{2}}\] \[mo{{l}^{-1}}\]
B) 1, 24.62 \[d{{m}^{3}}\]atm \[mo{{l}^{-1}},\]\[1.65\times {{10}^{-3}}\]\[d{{m}^{-6}}\]\[mo{{l}^{-2}}\]
C) \[1,4.1\times {{10}^{-2}}d{{m}^{-3}}\]\[at{{m}^{-1}}\]mol, \[606.0\,d{{m}^{6}}\]\[at{{m}^{2}}\]\[mo{{l}^{-2}}\]
D) \[24.62\,d{{m}^{3}}\]atm \[mo{{l}^{-1}},\] \[606.\,d{{m}^{6}}\] \[at{{m}^{2}}\]\[at{{m}^{2}}\]\[mo{{l}^{-2}}\],\[1.65\times {{10}^{-3}}\] \[d{{m}^{-6}}\] \[at{{m}^{-2}}\]\[mo{{l}^{2}}\]
Correct Answer: B
Solution :
\[{{K}_{P}}={{K}_{C}}{{(RT)}^{\Delta ng}}\]
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