Chemical EquilibriumMedium
Question
PCl5(g) $\rightleftharpoons$PCl3(g) + Cl2(g). In the above reaction, the partial pressure of PCl3, Cl2 and PCl5 are 0.3, 0.2 and 0.6 atm, respectively. If partial pressure of PCl3 and Cl2 was increased twice at the new equilibrium, then what will be the new partial pressure of PCl5 (in atm)?
Options
A.0.3
B.1.2
C.2.4
D.0.15
Solution
Kp =$\frac{P_{PCl_{3}}.P_{Cl_{2}}}{P_{PCl_{5}}}$= Constant
If $P_{PCl_{3}}' = 2 \times P_{PCl_{3}}$ and $P_{Cl_{2}}' = 2 \times P_{Cl_{2}}$ then $P_{PCl_{5}}' = 4 \times P_{PCl_{5}}$.
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