The H 2 S gas left in the container is now completely decomposed at the same temperature by 2 H 2 S ( g ) → 2 H 2 ( g ) + S 2 ( g ) . Find the new height difference of the liquid levels in the manometer.

3 mol H 2 S gives 3 mol H 2 + 1.5 mol S 2 = 4.5 mol gas. Left side: P = ( 4.5 × 22.4 ) /67.2 = 1.5 atm . Right side stays at 0.5 atm, so Δ P = 1.0 atm = 76 cm Hg . h = 76 × 2 = 152 cm . Answer: (C).

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The figure shows the initial conditions of a system when all valves are closed.

At $0^{\circ} C$ all valves are opened. Take the reaction $2 H_{2} S (g) + SO_{2} (g) \to 3 S (s) + 2 H_{2} O (l)$ to go to completion.

Assume the manometer is sufficiently long and thin. Neglect the volume of solids and liquid formed, the vapour pressure of water, and the volume of connecting tubes.

The $H_{2} S$ gas left in the container is now completely decomposed at the same temperature by $2 H_{2} S (g) \to 2 H_{2} (g) + S_{2} (g)$ . Find the new height difference of the liquid levels in the manometer.

Options:

Answer:

Solution:

3 mol $H_{2} S$ gives 3 mol $H_{2}$ + 1.5 mol $S_{2}$ $= 4.5$ mol gas. Left side: $P = (4.5 \times 22.4) /67.2 = 1.5 atm$ .

Right side stays at 0.5 atm, so $Δ P = 1.0 atm = 76 cm Hg$ . $h = 76 \times 2 = 152 cm$ . Answer: (C).

Stream:JEESubject:ChemistryTopic:Ideal GasSubtopic:Problems related with container/piston/SPM

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