PCls (g) PCl3 (8) + Cl2 (8) When 0.30 moles of PCl5 (g) are removed from the equ
ID: 1088460 • Letter: P
Question
PCls (g) PCl3 (8) + Cl2 (8) When 0.30 moles of PCl5 (g) are removed from the equilibrium system at constant temperature: the value of KcA. increases B. decreases. C. remains the same. the value of QcA. is greater than Ke B. is equal to Kc C. is less than Kc. A. run in the forward direction to reestablish equilibrium. B. run in the reverse direction to reestablish equilibrium. C. remain the same. It is already at equilibrium. the reaction must: the concentration of PC, will: A. increase. B. decrease. C. remain the same.Explanation / Answer
For the reaction at equilibrium,
PCl5(g) <==> PCl3(g) + Cl2(g)
when 0.30 mol of PCl5(g) is removed,
Kc : C. remains the same [This is a constant for a particular reaction]
Qc : A. Is greater than Kc [As subtrate concentrate decreases, product/substrate ratio increases]
the reaction must : B. run in the reverse direction to reestablish equilibrium
the concentration of PCl3 will : B. decrease [until it equals the ratio needed for Kc]
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