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Q.

In a closed container and at constant temperature 0.3 mole of SO2 and 0.2 mole of O2 gas at 750 torr are kept with a catalyst. If at equilibrium 0.2 mole of SO3 is formed the partial pressure of SO2 is ........ torr

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a

150

b

360

c

187

d

375

answer is D.

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Detailed Solution

Initial moles of SO2 = 0.3

Initial moles of O2 = 0.2

moles of SO3 at equilibrium = 0.2

2 mole of SO2 gives 2 moles of SO3

_____                          0.2 moles of SO3

moles of SO2 reacted = 0.2

1 mole of O2 gives 2 moles of SO3

_____             gives 0.2 moles of SO3

  \large \,2S{O_2}+ \large O_2 \large \rightleftharpoons \large 2S{O_3}\,
Initial moles 0.3 0.2   _____
Moles at equilibrium (0.3 - 0.2) (0.2 - 0.1)   0.2
      ni = (0.3 + 0.2) = 0.5 neq = 0.4
      Pi = 750 torr Peq = ?

\large \frac{{{P_i}}}{{{n_i}}} = \frac{{{P_{eq}}}}{{{n_{eq}}}}

\large \frac{{750}}{{0.5}} = \frac{{{P_{eq}}}}{{0.4}}

\large \boxed{{P_{eq}} = 600\, torr}

\large {P_{S{O_2}}} = {X_{S{O_2}}} \times {\text{Total pressure}}

\large {P_{S{O_2}}} = \frac{{0.1}}{{0.4}} \times 600

\large \boxed{{P_{S{O_2}}} = 150\,torr}

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