A) \[p={{K}_{p}}\]
B) \[p=3{{K}_{p}}\]
C) \[p=4{{K}_{p}}\]
D) \[p=8{{K}_{p}}\]
Correct Answer: D
Solution :
\[AB(g)A(g)+B(g)\] 1 0 0 Initially 1 - 0.33 0.33 0.33 When 33% dissociated = 0.67 Total pressure at equilibrium, = 1 - 0.33 + 0.33 + 0.33 = 1 + 0.33 = 1.33 \[{{p}_{A}}=\frac{0.33}{1.33}p\] \[{{p}_{B}}=\frac{0.33}{1.33}p\] \[{{p}_{AB}}=\frac{0.67}{1.33}p\] (Where, \[{{p}_{A}},{{p}_{B}}\] and \[{{p}_{A}}_{B}\] are partial pressures of A, B and AB.) \[{{k}_{p}}=\frac{{{p}_{A}}.\,{{p}_{B}}}{{{p}_{AB}}}\] \[=\frac{\frac{0.33}{1.33}p\,.\,\,\frac{0.33}{1.33}p}{\frac{0.67}{1.33}p}\] \[=\frac{{{(0.33)}^{2}}p}{1.33\times 0.67}\] \[=\frac{01089\times p}{0.8911}\] \[{{k}_{p}}=0.122p\] \[p=\frac{{{K}_{p}}}{0.122}=8{{K}_{p}}\]
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