Chemical EquilibriumHard
Question
Under what pressure must an equimolar mixture of Cl2 and PCl3 be placed at 250oC in order to obtain 75% conversion of PCl3 into PCl5? Given that PCl3(g) + Cl2(g) $\rightleftharpoons$PCl5(g); KP = 2 atm–1.
Options
A.12 atm
B.6 atm
C.15 atm
D.30 atm
Solution
PCl2 (g) + Cl2 (g) $\rightleftharpoons$ PCl5 (g)
Initial partial pressure P0 P0 0
Equilibrium partial pressure P0 – 0.75 P0 P0 – 0.75 P0 0.75 P0
–0.25 P0 –0.25 P0
Now, $K_{p} = \frac{P_{PCl_{5}}}{P_{PCl_{3}} \times P_{Cl_{2}}}$
$\Rightarrow 2 = \frac{0.75P_{o}}{0.25P_{o} \times 0.25P_{o}} \Rightarrow P_{o} = 6\text{ atm}$
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