| The equilibrium constants at 1140 K for these two reactions are |
| \[2CO(g)C(s)+C{{O}_{2}}(g);{{K}_{p1}}={{10}^{-12}}at{{m}^{-1}}\] |
| \[CO(g)+C{{l}_{2}}(g)COC{{l}_{2}}(g);{{K}_{p2}}=3\times {{10}^{-3}}\text{at}{{\text{m}}^{-1}}\] |
| What will be equilibrium constant \[{{K}_{c}}\] for the following reaction at 1140 K? |
| \[C(s)+C{{O}_{2}}(g)+2C{{l}_{2}}(g)2COC{{l}_{2}}(g)\] |
A) \[8.42\times {{10}^{8}}{{M}^{-1}}\]
B) \[8.42\times {{10}^{4}}{{M}^{-1}}\]
C) \[4.21\times {{10}^{4}}{{M}^{-1}}\]
D) \[4.21\times {{10}^{8}}{{M}^{-1}}\]
Correct Answer: A
Solution :
| [a] \[C(s)+C{{O}_{2}}(g)2CO(g)\] ?(i) |
| \[K{{'}_{{{p}_{1}}}}={{10}^{12}}atm,K{{'}_{{{p}_{1}}}}\frac{1}{{{K}_{{{p}_{1}}}}}=\frac{1}{{{10}^{-12}}}\] |
| \[2CO(g)+2C{{l}_{2}}(g)2COC{{l}_{2}}(g)\] ?(ii) |
| \[{{K}_{p{{'}_{2}}}}={{(3\times {{10}^{-3}})}^{2}}\text{at}{{\text{m}}^{\text{-2}}}{{K}_{p{{'}_{2}}}}={{({{K}_{{{p}_{2}}}})}^{2}}\] |
| Adding Egs. (i) and (ii), we get |
| \[C(s)+C{{O}_{2}}(g)+2C{{l}_{2}}(g)2COC{{l}_{2}}(g)\] |
| \[{{K}_{p}}={{10}^{12}}\times 9\times {{10}^{-6}}=9\times {{10}^{6}}\] |
| \[\Delta ng=2-(1+2)=-1,{{K}_{p}}={{K}_{c}}{{(RT)}^{-1}}\] |
| \[{{K}_{c}}={{K}_{p}}(RT)\] |
| \[{{K}_{c}}=9\times {{10}^{6}}\times 0.0821\times 1140\] |
| \[=8.42\times {{10}^{8}}{{M}^{-1}}\] |
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