| \[{{N}_{2}}\left( g \right)+{{O}_{2}}\left( g \right)\rightleftharpoons 2\text{ }NO\left( g \right)\] |
| \[{{N}_{2}}{{O}_{4}}(g)\,\,\rightleftharpoons \,\,2NO\,(g)\] |
| \[{{N}_{2}}(g)+3{{H}_{2}}(g)\rightleftharpoons 2N{{H}_{3}}(g)\] |
A) 24.62 \[d{{m}^{3}}\] atm \[mo{{l}^{-1}}\], 606.0 \[d{{m}^{6}}\] \[at{{m}^{2}}\]\[mo{{l}^{-2}}\] \[1.65\times {{10}^{-3}}d{{m}^{-}}^{6}at{{m}^{-}}^{2}mo{{l}^{2}}\]
B) \[1,4.1\times {{10}^{-}}^{2}d{{m}^{-}}^{3}\,at{{m}^{-}}^{1}\] mol, 606 \[d{{m}^{6}}\] \[at{{m}^{2}}\,mo{{l}^{-}}^{2}\]
C) \[1,24.62\text{ }d{{m}^{3}}\text{ }atm\text{ }mo{{l}^{-}}^{1}\], \[1.65\times {{10}^{-3}}\] \[d{{m}^{-6}}\]\[at{{m}^{-}}^{2}\,mo{{l}^{2}}\]
D) 1, 24.62 \[d{{m}^{3}}\] atm\[mo{{l}^{-}}^{1}\], 606.0 \[d{{m}^{6}}\] \[at{{m}^{2}}\]\[mo{{l}^{-}}^{2}\]
Correct Answer: C
Solution :
\[\frac{{{K}_{P}}}{{{K}_{C}}}{{(RT)}^{{{\Delta }_{n}}_{_{g}}}}\] for reaction I, \[\Delta {{n}_{g}}=0\] \[\therefore \,\,\,{{K}_{P}}={{K}_{C}}\Rightarrow \frac{{{K}_{P}}}{{{K}_{C}}}=1\] for reaction II, \[\Delta {{n}_{g}}=1\] \[\therefore \,\,\text{ }\frac{{{K}_{P}}}{{{K}_{C}}}={{\left( RT \right)}^{1}}=24.62\text{ }d{{m}^{3}}atm\text{ }mo{{l}^{-1}}\] for reaction III, \[\Delta {{n}_{g}}=\left( -2 \right)\] \[\therefore \,\,\text{ }\frac{{{K}_{P}}}{{{K}_{C}}}={{\left( RT \right)}^{-2}}\] \[=1.649\times {{10}^{-3}}\,d{{m}^{-6}}\,at{{m}^{-2}}\,mo{{\text{l}}^{\text{2}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
