$2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{SO}_{3}(\mathrm{~g})$ The above reaction is carried out in a vessel starting with partial pressure $\mathrm{P}_{\mathrm{SO}_{2}}=250 \mathrm{~m}$ bar, $\mathrm{P}_{\mathrm{O}_{2}}=750 \mathrm{~m}$ bar and $\mathrm{P}_{\mathrm{SO}_{3}}=0$ bar. When the reaction is complete, the total pressure in the reaction vessel is $\_\_\_\_$ m bar. (Round off to the Nearest Integer).

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · States of Matter

2 SO_{2} (g) + O_{2} (g) \to 2 SO_{3} (g)

The above reaction is carried out in a vessel starting with partial pressure

P_{SO_{2}} = 250 m

bar,

P_{O_{2}} = 750 m

bar and

P_{SO_{3}} = 0

bar. When the reaction is complete, the total pressure in the reaction vessel is

____

m bar.
(Round off to the Nearest Integer).

Answer: 875

Solution

The equation needs

S O_{2}

and

O_{2}

in the ratio

2 : 1

, so

250

m bar of

S O_{2}

can consume only

125

m bar of

O_{2}

. With

750

m bar of oxygen available,

S O_{2}

is the limiting reactant and disappears completely, producing an equal pressure of

S O_{3}

P_{S O_{2}} : 250 \to 0, P_{O_{2}} : 750 \to 625, P_{S O_{3}} : 0 \to 250

Adding the partial pressures remaining at completion,

0 + 625 + 250 = 875

, so the total pressure in the vessel is

875

m bar.

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Integer answer

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