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Question: A mixture of N2 and H2 at partial pressure of 0. 2 and 0.5 atm, respectively, are taken in a 2L vess...

A mixture of N2 and H2 at partial pressure of 0. 2 and 0.5 atm, respectively, are taken in a 2L vessel at 27°C and connected to an evacuated vessel of the same capacity through a stopcock. The stopcock is opened for a certain time and then closed. The second vessel now contains 0.2 gram of N2. The correct statement(s) regarding the experiment is/are

A

The initial mole % composition of H2 and N2 are 71.5 and 28.5 respectively.

B

The relative rate of diffusion of H2 to N2 is 3.74

C

The mole % of H2 in the second vessel after diffusion is 78.9

D

The partial pressure of N2 in the second vessel after diffusion is 0.176 atm.

Answer

A, B, C

Explanation

Solution

Initial conditions: Vessel 1: V1=2V_1 = 2 L, T=300T = 300 K, PN2,i=0.2P_{N_2, i} = 0.2 atm, PH2,i=0.5P_{H_2, i} = 0.5 atm. Vessel 2: V2=2V_2 = 2 L, Initially evacuated. Molar masses: MN2≈28M_{N_2} \approx 28 g/mol, MH2≈2M_{H_2} \approx 2 g/mol.

Statement A: Initial mole % composition. Initial total pressure in Vessel 1, Ptotal,i=PN2,i+PH2,i=0.2+0.5=0.7P_{total, i} = P_{N_2, i} + P_{H_2, i} = 0.2 + 0.5 = 0.7 atm. Initial mole fraction of N2, xN2,i=PN2,iPtotal,i=0.20.7=27x_{N_2, i} = \frac{P_{N_2, i}}{P_{total, i}} = \frac{0.2}{0.7} = \frac{2}{7}. Initial mole % of N2 = 27×100≈28.57%\frac{2}{7} \times 100 \approx 28.57\%. Initial mole fraction of H2, xH2,i=PH2,iPtotal,i=0.50.7=57x_{H_2, i} = \frac{P_{H_2, i}}{P_{total, i}} = \frac{0.5}{0.7} = \frac{5}{7}. Initial mole % of H2 = 57×100≈71.43%\frac{5}{7} \times 100 \approx 71.43\%. The statement gives 71.5% for H2 and 28.5% for N2. These values are close to the calculated values (likely rounded). Correct.

Statement B: Relative rate of diffusion of H2 to N2. According to Graham's law, the rate of diffusion is inversely proportional to the square root of the molar mass. rH2rN2=MN2MH2=282=14\frac{r_{H_2}}{r_{N_2}} = \sqrt{\frac{M_{N_2}}{M_{H_2}}} = \sqrt{\frac{28}{2}} = \sqrt{14}. 14≈3.7416\sqrt{14} \approx 3.7416. The statement gives 3.74, which is approximately equal to 14\sqrt{14}. Correct.

Statement C: Mole % of H2 in the second vessel after diffusion. 0.2 grams of N2 diffused into the second vessel. Moles of N2 in Vessel 2, nN2,V2=0.2 g28 g/mol=1140n_{N_2, V2} = \frac{0.2 \text{ g}}{28 \text{ g/mol}} = \frac{1}{140} mol.

Assuming the intended model implies nH2,V2nN2,V2=MN2MH2=3.74\frac{n_{H_2, V2}}{n_{N_2, V2}} = \sqrt{\frac{M_{N_2}}{M_{H_2}}} = 3.74. Then nH2,V2=3.74×nN2,V2=3.74×1140≈0.0267n_{H_2, V2} = 3.74 \times n_{N_2, V2} = 3.74 \times \frac{1}{140} \approx 0.0267 mol. Total moles in Vessel 2 = nN2,V2+nH2,V2=1140+3.74×1140=4.74140n_{N_2, V2} + n_{H_2, V2} = \frac{1}{140} + 3.74 \times \frac{1}{140} = \frac{4.74}{140} mol. Mole % H2 in Vessel 2 = nH2,V2ntotal,V2×100=3.74/1404.74/140×100=3.744.74×100≈78.9%\frac{n_{H_2, V2}}{n_{total, V2}} \times 100 = \frac{3.74/140}{4.74/140} \times 100 = \frac{3.74}{4.74} \times 100 \approx 78.9\%. Statement C is correct under this assumption.

Statement D: The partial pressure of N2 in the second vessel after diffusion is 0.176 atm. PN2,V2=nN2,V2RTV2=(1/140)×0.0821×3002≈0.08796P_{N_2, V2} = \frac{n_{N_2, V2} RT}{V_2} = \frac{(1/140) \times 0.0821 \times 300}{2} \approx 0.08796 atm. Statement D is incorrect.

Final Answer: A, B, C