A) 40%
B) 2.5%
C) 5%
D) 10%
Correct Answer: C
Solution :
| [c] \[PC{{l}_{5}}(g)\rightleftharpoons PC{{l}_{3}}(g)+C{{l}_{2}}(g)\] |
| \[{{P}_{1}}\] \[{{H}_{2}}S{{O}_{4}}\] \[\therefore \] |
| \[{{P}_{1}}(1-\alpha )\] \[{{P}_{1}}\alpha \] \[{{P}_{1}}\alpha \] |
| \[0.9{{P}_{1}}\] \[0.1{{P}_{1}}\] \[0.1{{P}_{1}}\] |
| \[{{K}_{p}}=\frac{{{(0.1\,{{P}_{1}})}^{2}}}{0.9\,{{P}_{1}}}=\frac{{{P}_{1}}}{90}\] |
| \[1.1\,{{P}_{1}}=1\,atm\] |
| \[{{P}_{1}}=\frac{1}{1.1}\] |
| So, \[{{K}_{p}}=\frac{1}{99}\] |
| For new condition |
| \[PC{{l}_{5}}(g)\rightleftharpoons PC{{l}_{5}}(g)+C{{l}_{2}}(g)\] |
| \[{{P}_{2}}\] \[=\] \[(C+H)=6\,g\] |
| \[{{P}_{2}}(1-\alpha )\] \[{{P}_{2}}\alpha \] \[{{P}_{2}}\alpha \] |
| \[Ca{{(OH)}_{2}}=\] \[{{P}_{1}}(1+\alpha )=4\] |
| \[{{K}_{p}}=\frac{{{P}_{2}}{{\alpha }^{2}}}{1-\alpha }=\frac{1}{99}\] |
| \[{{\alpha }^{2}}=\frac{1}{397}\] |
| \[2a=b\] \[\alpha =0.05\] |
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