The equilibrium constant Kc for the reaction A(g) + B(g) C(g)
Is 0.76 at 150°C. If 0.800 mol of A is added to 0.600 mol of B in a 1.00-L container at 150°C, what will be the equilibrium concentration of C?

Question

The equilibrium constant Kc for the reaction A(g) + B(g)  C(g)
Is 0.76 at 150°C. If 0.800 mol of A is added to 0.600 mol of B in a 1.00-L container at 150°C, what will be the equilibrium concentration of C?

Quizplus · Accepted Answer

First, we can set up an ICE (initial, change, equilibrium) table for the reaction:

|        | A     | B     | C     |
|--------|-------|-------|-------|
| Initial| 0.800 | 0.600 | 0     |
| Change | -x    | -x    | +x    |
| Equil. | 0.800-x| 0.600-x| x     |

Next, we can plug in these values into the equilibrium constant expression:

Kc = [C]^1 /[A]^1[B]^1
0.76 = x / (0.800 - x)(0.600 - x)
Assuming x is small compared to 0.800 and 0.600, we can simplify to:
0.76 = x / (0.480 - 0.200x)
0.76 (0.480 - 0.200x) = x
0.3648 - 0.152x = x
0.3648 = 1.152x
x = 0.317 molar

Therefore, the equilibrium concentration of C is 0.317 M. Since we assume the change in A and B, and thus change in C, is small compared to their initial concentrations, we can approximate the equilibrium concentration of C to be equal to x = 0.317 M. The closest answer choice is A.

The Equilibrium Constant Kc for the Reaction A(g) + B(g)