Using the Equilibrium Constant

Question

The reversible chemical reaction A+B<=>C+D has the following equilibrium constant: Kc=[C][D]/[A][B]=2.4.

 

Part A

Initially, only A and B are present, each at 2.00 M. What is the final concentration of A once equilibrium is reached?

 

Part B

What is the final concentration of D at equilibrium if the initial concentrations are [A]=1.00 M and [B]=2.00 M?

Answer 1

Part A:

Setup an ICE table. Initial concentration of A,B,C, D are 2,2,0, and 0, respectively; The change in concentrations are -x, -x, +x, +x of A,B,C, and D; and the final concentrations are 2-x,2-x,x,x for A,B,C,D. Then put these values in for the equilibrium constant equation to get x²/(4-4x-x²) = 2.4. Solve for x using the quadratic formula. To get the final concentration of A, solve 2-x.

 

Part B:

Once again, etup an ICE table. Initial concentration of A,B,C, D are 1,2,0, and 0, respectively; The change in concentrations are -x, -x, +x, +x of A,B,C, and D; and the final concentrations are 1-x,2-x,x,x for A,B,C,D. Then put these values in for the equilibrium constant equation to get x²/(2-2x-x²) = 2.4. Solve for x using the quadratic formula. The final concentration of D is x.