Question

The reaction ${ 2A_{(g)} + B_{(g)} <=> 3C_{(g)} + D_{(g)}}$ is begun with the concentrations of $A$ and $B$ both at an intial 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. $[(0.75)^3 (0.25)] \div [(0.50)^2 (0.75)]$
• B. $[(0.75)^3 (0.25)] \div [(0.50)^2 (0.25)]$
• C. $[(0.75)^3 (0.25)] \div [(0.75)^2 (0.25)]$
• D. $[(0.75)^3 (0.25)] \div [(1.00)^2 (1.00)]$

Answer 1

The Correct Option is A

To solve this problem, we need to find the equilibrium constant expression based on the initial and equilibrium concentrations of the reactants and products in the balanced chemical equation. The chemical reaction provided is:

{ 2A_{(g)} + B_{(g)} <=> 3C_{(g)} + D_{(g)}}

We are given the initial concentrations of A and B are both 1.00\, M, and the equilibrium concentration of D is 0.25 \,M.

1. Let's set up the equilibrium concentration expression based on the stoichiometry of the reaction:
   At equilibrium, the change in concentration for each species compared to initial concentrations can be expressed in terms of x:
   • The change in concentration of B would be -x.
   • The change in concentration of A would be -2x (because for each 2 moles of A, reaction produces 3 moles of C).
   • The increase in concentration of C would be +3x.
   • The increase in concentration of D would be +x.
2. Using [D] = 0.25 \, M, we deduce that x = 0.25.
3. Calculate concentrations at equilibrium:
   • [A] = 1.00 - 2 \times 0.25 = 0.50 \, M
   • [B] = 1.00 - 0.25 = 0.75 \, M
   • [C] = 3 \times 0.25 = 0.75 \, M
   • [D] = 0.25 \, M (as given)
4. The equilibrium constant expression for this reaction is given by: K_c = \dfrac{[C]^3 [D]}{[A]^2 [B]}
5. Substitute the equilibrium concentrations into the expression:
   K_c = \dfrac{(0.75)^3 (0.25)}{(0.50)^2 (0.75)}

The correct answer is therefore: [(0.75)^3 (0.25)] \div [(0.50)^2 (0.75)].