Q. The reaction 2A(g) + B(g) ⇌ 3C(g) + D(g) is begun with the concentrations of A and B both at an initial value of 1.00 M. When equilibrium is reached, the concentration of D is measured and found to be 0.25 M. The value for the equilibrium constant for this reaction is given by the expression
A) [(0.75)³(0.25)] ÷ [(1.00)²(1.00)]
B) [(0.75)³(0.25)] ÷ [(0.50)²(0.75)]
C) [(0.75)³(0.25)] ÷ [(0.50)²(0.25)]
D) [(0.75)³(0.25)] ÷ [(0.75)²(0.25)]
Explanation
[Solution from source] Forming 0.25 M of D consumes 0.25 M B and 0.50 M A and forms 0.75 M C. At equilibrium [A] = 0.50, [B] = 0.75, [C] = 0.75, [D] = 0.25, so K = (0.75)³(0.25)/[(0.50)²(0.75)]. Using the initial 1.00 M values in the denominator ignores the amounts consumed. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.