The values of the Henry's law constant $A r, C O_{2}, C H_{4} a n d O_{2}$ in water at $25^{o} C$ are 40.30, 1.67, 0.41 and 34.86 kbar respectively. The order of their solubility in water at the same temperature and pressure is

A r > O_{2} > C O_{2} > C H_{4}

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C H_{4} > C O_{2} > A r > O_{2}

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C H_{4} > C O_{2} > O_{2} > A r

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A r > C H_{4} > O_{2} > C O_{2}

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Solution

The correct option is C $C H_{4} > C O_{2} > O_{2} > A r$
According to Henry's law,
$P_{g a s} = K_{H} \times X_{g a s}$
$Xgas(molefraction)∝solubility of gas∝1KH$
$K_{H}$ value for $A r, C O_{2}, C H_{4} a n d O_{2}$ are 40.30, 1.67, 0.41, 34.86 so solubility order is
$C H_{4} > C O_{2} > O_{2} > A r$

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Question 26
$K_{H}$ value for Ar (g), $C O_{2}$ (g), HCHO (g) and $C H_{4}$ (g) are 40.39, 1.67, $1.83 \times 10^{- 5}$ and 0.413 repectively.
Arrange these gases in the order of their increasing solubility.
(a) $H C H O < C H_{4} < C O_{2} < A r$ (b) $H C H O < C O_{2} < C H_{4} < A r$
(c) $A r < C O_{2} < C H_{4} < H C H O$ (d) $A r < C H_{4} < C O_{2} < H C H O$

Q. What coefficients are required to balance the following equation?

{C H}_{4} + O_{2} \to {C O}_{2} + H_{2} O

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The values of the Henry's law constant Ar, CO2, CH4 and O2 in water at 25 oC are 40.30, 1.67, 0.41 and 34.86 kbar respectively. The order of their solubility in water at the same temperature and pressure is

The correct option is C CH4>CO2>O2>ArAccording to Henry's law, Pgas=KH×Xgas Xgas(molefraction)∝solubility of gas∝1KH KH value for Ar,CO2,CH4 and O2 are 40.30, 1.67, 0.41, 34.86 so solubility order is CH4>CO2>O2>Ar

The values of the Henry's law constant $A r, C O_{2}, C H_{4} a n d O_{2}$ in water at $25^{o} C$ are 40.30, 1.67, 0.41 and 34.86 kbar respectively. The order of their solubility in water at the same temperature and pressure is