Question

Equilibrium is established for the reaction 2 X(s) + Y(g) ⇌ 2 Z(g) at 500K, the Kc value is 100.

Determine the concentration of Z in equilibrium with o.2 mol of X and 0.50 M of Y at 500K

Answer 1

The concentration of Z in equilibrium can be determined using the equilibrium constant (Kc) and the initial concentrations of X and Y: [X] = 0.2 mol, [Y] = 0.50 M, [Z] = 0. The equilibrium concentration of Z is 0.2 M.

Step-by-step explanation:

The concentration of Z in equilibrium can be determined using the equilibrium constant (Kc) and the initial concentrations of X and Y. The given Kc value of 100 indicates that the reaction favors the formation of Z. To determine the concentration of Z, we can set up an ICE table:

1. Initial: [X] = 0.2 mol, [Y] = 0.50 M, [Z] = 0
2. Change: -2x, -x, +2x
3. Equilibrium: [X] = 0.2 - 2x, [Y] = 0.50 - x, [Z] = 2x

Since the equilibrium concentration of X is given as 0.2 mol, we can substitute that into the equation:

0.2 - 2x = 0

Solving for x, we find x = 0.1 mol. Therefore, the equilibrium concentration of Z is 2x = 2(0.1) = 0.2 M.

Answer 2

The concentration of Z in equilibrium with 0.2 mol of X and 0.50 M of Y at 500K is approximately √2 M.

Step-by-step explanation:

Equilibrium can be represented by the equation:

2 X(s) + Y(g) ⇌ 2 Z(g)

At 500K, the equilibrium constant (Kc) is given as 100.

To determine the concentration of Z in equilibrium with 0.2 mol of X and 0.50 M of Y at 500K, we can use the expression for Kc:

Kc = [Z]² / ([X]² [Y])

Rearranging the equation, we get:

[Z]² = Kc * [X]² [Y]

Substituting the given values, we have:

[Z]² = 100 * (0.2 mol)² (0.50 M)

[Z]² = 2

Therefore, the concentration of Z in equilibrium is √2 M.