For the reaction given below,2.00 moles of A and 3.00 moles of B are placed in a 6.00-L container.
A(g) + 2B(g) C(g)
At equilibrium,the concentration of A is 0.223 mol/L.What is the value of K_c?

Question

For the reaction given below,2.00 moles of A and 3.00 moles of B are placed in a 6.00-L container. A(g)+ 2B(g) C(g) At equilibrium,the concentration of A is 0.223 mol/L.What is the value of K_c?

Quizplus · Accepted Answer

We can start by setting up an ICE table:

|             | A        | B        | C        |
| ----------- | -------- | -------- | -------- |
| Initial     | 2.00 mol | 3.00 mol | 0 mol    |
| Change      | -x       | -2x      | +x       |
| Equilibrium | 2.00-x   | 3.00-2x  | x        |

The equilibrium constant expression for the reaction is:

Kc = [C]^1/[A]^1[B]^2

Plugging in the values from the ICE table, we get:

Kc = x/[2.00-x][3.00-2x]^2

Since we are given that the concentration of A at equilibrium is 0.223 mol/L, we can substitute 0.223 for [A]:

0.223 = 2.00-x

Solving for x, we get:

x = 1.777 mol/L

Substituting this value back into the equilibrium constant expression, we get:

Kc = 1.777 / [0.223]^1 [3.00-2(1.777)]^2 = 6.34

Therefore, the answer is C.

For the Reaction Given Below,2