Chemical EquilibriumMedium
Question
In a closed container maintained at 1 atm pressure and 25o C, 2 moles of SO2(g) and 1 mole of O2(g) were allowed to react to form SO3(g) under the influence of a catalyst. 2SO2(g) + O2(g)$\rightleftharpoons$ 2SO3(g)
At equilibrium, it was found that 50% of SO2(g) was converted to SO3(g). The partial pressure of O2(g) at equilibrium will be
Options
A.0.17 atm
B.0.5 atm
C.0.33 atm
D.0.20 atm
Solution
2SO2 + O2 $\rightleftharpoons$ 2SO3
2 mole 1 mole 0
Equilibrium $2 - 2 \times \frac{50}{100} = 1$ $1 - \frac{1}{2} = 0.5$
$\therefore P_{O_{2}} = \frac{\frac{1}{2} \times 0.5}{2.5} \times 1 = 0.2\text{ atm}$
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