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A mixture of 0.5 mol of CH4, 0.5 mol of H2, and 0.5 mol of SO2 is introduced into a 10.0 L container at 25 *C.

If the container has a pinhole leak, which describes the relationship between the partial pressures of the individual components in the containtainer after 3 hours?

(A.) PSO2 > PCH4>PH2.

(B.) PSO2
(C) PSO2 < PCH4 > PH2.

(D) PSO2= PCH4 = PH2.

Respuesta :

Answer: The partial pressure of individual components in the container will be same, that is [tex]p_{SO_2}=p_{CH_4}=p_{H_2}[/tex]

Explanation:

To calculate the total pressure, we use the equation given by ideal gas equation:

[tex]PV=nRT[/tex]

where,

P = pressure of the gas

V = Volume of gas = 10.0 L

n = Number of moles = [tex](n_{CH_4}+n_{H_2}+n_{SO_2})=(0.5+0.5+0.5)=1.5mol[/tex]

R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]

T = temperature of the gas = [tex]25^oC=[25+273]K=298K[/tex]

Putting values in above equation, we get:

[tex]P\times 10L=1.5\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\\\P=\frac{1.5\times 0.0821\times 298}{10}=3.67atm[/tex]

The partial pressure of a gas is given by Raoult's law, which is:

[tex]p_A=p_T\times \chi_A[/tex]     ......(1)

where,

[tex]p_A[/tex] = partial pressure of substance A

[tex]p_T[/tex] = total pressure

[tex]\chi_A[/tex] = mole fraction of substance A

To calculate the mole fraction of a substance, we use the equation:

[tex]\chi_A=\frac{n_{A}}{n_{A}+n_B}[/tex]       .......(2)

We are given:

Moles of methane = 0.5 moles

Moles of hydrogen = 0.5 moles

Moles of sulfur dioxide = 0.5 moles

  • For methane:

Using equation 2, we get:

[tex]\chi_{CH_4}=\frac{n_{CH_4}}{n_{CH_4}+n_{H_2}+n_{SO_2}}=\frac{0.5}{1.5}=\frac{1}{3}[/tex]

Using equation 1, we get:

[tex]p_{CH_4}=3.67\times \frac{1}{3}=1.22atm[/tex]

Partial pressure of methane = 1.22 atm

  • For hydrogen gas:

Using equation 2, we get:

[tex]\chi_{H_2}=\frac{n_{H_2}}{n_{CH_4}+n_{H_2}+n_{SO_2}}=\frac{0.5}{1.5}=\frac{1}{3}[/tex]

Using equation 1, we get:

[tex]p_{H_2}=3.67\times \frac{1}{3}=1.22atm[/tex]

Partial pressure of hydrogen gas = 1.22 atm

  • For sulfur dioxide:

Using equation 2, we get:

[tex]\chi_{SO_2}=\frac{n_{SO_2}}{n_{CH_4}+n_{H_2}+n_{SO_2}}=\frac{0.5}{1.5}=\frac{1}{3}[/tex]

Using equation 1, we get:

[tex]p_{SO_2}=3.67\times \frac{1}{3}=1.22atm[/tex]

Partial pressure of sulfur dioxide = 1.22 atm

Hence, the partial pressure of individual components in the container will be same, that is [tex]p_{SO_2}=p_{CH_4}=p_{H_2}[/tex]

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