A) \[\frac{1}{2}\]
B) \[\frac{2}{1}\]
C) \[\frac{1}{4}\]
D) \[\frac{4}{1}\]
Correct Answer: D
Solution :
[d] \[{{N}_{2}}{{O}_{4}}\rightleftharpoons 2N{{O}_{2}}\] If \[\alpha \] is the degree of dissociation and P is the total pressure, then \[{{K}_{p}}=\frac{4{{\alpha }^{2}}}{1-{{\alpha }^{2}}}P\] when \[\alpha \] =0.10, \[{{K}_{p}}=\frac{4{{(0.10)}^{2}}}{1-{{(0.10)}^{2}}}\times {{P}_{1}}=\frac{0.04}{0.99}{{P}_{1}}\simeq 0.04{{P}_{1}}\] When \[\alpha \]=0.20, \[{{K}_{p}}=\frac{4{{(0.20)}^{2}}}{1-{{(0.2)}^{2}}}\times {{P}_{2}}=\frac{0.16}{0.96}{{P}_{2}}\simeq 0.16{{P}_{2}}\] Thus, \[0.04{{P}_{1}}=0.16{{P}_{2}}\]or \[\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{0.16}{0.96}=\frac{4}{1}\]
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