K_H is Henry's constant and has the unit | vedclass

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Question

(D) Henry's law is expressed as $P = K_H 	imes C$,where $P$ is the partial pressure of the gas and $C$ is the concentration of the gas in the solutio

Vedclass · Accepted Answer

$mol \ dm^{-3} \ atm^{-1}$

Vedclass · Answer

(D) Henry's law is expressed as $P = K_H 	imes C$,where $P$ is the partial pressure of the gas and $C$ is the concentration of the gas in the solution.Rearranging for the constant,we get $K_H = \frac{P}{C}$.The unit of pressure $P$ is $atm$ and the unit of concentration $C$ is $mol \ dm^{-3}$.Therefore,the unit of $K_H$ is $\frac{atm}{mol \ dm^{-3}} = atm \ dm^3 \ mol^{-1}$ or $mol^{-1} \ dm^3 \ atm$. However,in many textbooks,$K_H$ is expressed as $P = K_H 	imes x$ (where $x$ is mole fraction,unitless),making the unit of $K_H$ simply $atm$. If $K_H$ is defined as $C = K_H 	imes P$,then the unit is $mol \ dm^{-3} \ atm^{-1}$. Given the options,the correct unit for $K_H$ in the context of solubility concentration is $mol \ dm^{-3} \ atm^{-1}$.

Gas	$K_{H} / \text{kbar}$
$Ar$	$40.3$
$CO_{2}$	$1.67$
$HCHO$	$1.83 \times 10^{-5}$
$CH_{4}$	$0.413$

$K_H$ is Henry's constant and has the unit

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The $K_{H}$ values for some gases at the same temperature '$T$' are given below:

Gas $K_{H} / \text{kbar}$

$Ar$ $40.3$

$CO_{2}$ $1.67$

$HCHO$ $1.83 \times 10^{-5}$

$CH_{4}$ $0.413$

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