Question

At ${\mathbf{25}}^{\mathbf{0}}\mathit{C}$ , some benzene $\left({\text{C}}_6{\text{H}}_6\right)$is added to a sample of gascous methane $\left({\text{CH}}_4\right)$at 1.00 atmpressure in a closed vessel, and the vessel is shaken until as much methane as possible dissolves. Then 1.00 kg of the solution is removed and boiled to expel the methane, yielding the volume of methane that results is 0.510 L atrole="math" localid="1663672111295" ${\mathbf{0}}^{\mathbf{0}}\mathit{C}$ and 1.00 atm . Determine the Henry's law constant for methane in benzene.

Short answer

$5.66\times {10}^2\text{atm}$ is the constant for methane in benzene.

Step-by-step solution

Given data

At ${25}^{0}C$, water is added to methane $\left({\text{CH}}_4\right)$ at 1.00 atm pressure

1.00 kg of the solution is removed and boiled to expel the methane, yielding a volume of 3.01 L of ${\text{CH}}_4\left(g\right)$at $0^{0}C$ and 1.00 atm.

Concept of Henry’s law constant

Henry's Law constant relates the concentration of gas particles in the solution phase that is in equilibrium with the pressure of the gas in the vapor phase.First calculate the amount of methane dissolved in (and then, expelled from) the solvent sample. After that, the mole fraction can be expanded using the amounts - those will finally lead to the constant in question.

Calculate the amount of methane

Using the ideal gas law equation:

$$\begin{gathered} pV=nRT \\ (1.00\times 101.3)\times 0.510=n\times 8.3145\times (0+273.15) \\ n=0.0227\text{mol} \end{gathered}$$

Calculate the amount of benzene in the sample

Using the amount of methane in the molar mass formula calculates the amount of benzene.

$$\begin{gathered} n_{{\text{CH}}_4}=0.0227\text{mol} \\ M=16.0\text{g}/\text{mol}\backslash \mathrm{hfill} \\ {m}_{{\text{CH}}_4}=n\times M=0.364\text{g} \end{gathered}$$

$$\begin{gathered} M=78.0\text{g}/\text{mol} \\ {n}_{{\text{C}}_6{\text{H}}_6}=\frac{m}{M}=12.816\text{mol} \end{gathered}$$

Calculate the partial pressure of gas using Henry’s law

The law states that (at small mole fractions) the partial pressure of a gas above a solution is proportional to the dissolved part of it.

$$\begin{gathered} p_{{\text{CH}}_4}=k_{{\text{CH}}_4}\times X_{{\text{CH}}_4} \\ {p}_{{\text{CH}}_4}=k_{{\text{CH}}_4}\times \frac{n_{{\text{CH}}_4}}{n_{{\text{CH}}_4}+n_{{\text{C}}_6{\text{H}}_6}} \\ 1.00=k_{{\text{CH}}_4}\times \frac{0.0227}{0.0227+12.816} \\ {k}_{{\text{CH}}_4}=5.66\times {10}^2\text{atm} \end{gathered}$$

This pressure refers to the initial vessel containing pure methane gas at 1 atm .