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To solve the problem, we need to find the equilibrium constant K_c for the reaction given by:
X + Y \rightleftharpoons 2Z
### Step 1: Write the Expression for the Equilibrium Constant
The equilibrium constant K_c for the reaction can be expressed as:
K_c = \frac{[Z]^2}{[X][Y]}
### Step 2: Determine the Equilibrium Concentrations
From the problem, we know the initial amounts of the reactants and products:
- Initial moles of X = 1.0 \, \text{mol}
- Initial moles of Y = 1.5 \, \text{mol}
- Initial moles of Z = 0.5 \, \text{mol}
At equilibrium, the concentration of Z is given as 1.0 \, \text{mol L}^{-1} . Since the volume of the vessel is 1 \, \text{L} , this means:
[Z] = 1.0 \, \text{mol L}^{-1}
### Step 3: Calculate the Change in Concentration
Let x be the change in concentration of Z at equilibrium. Since 2Z is produced from X and Y , we can express the changes in concentrations as follows:
- For every mole of Z produced, \frac{1}{2} mole of X and \frac{1}{2} mole of Y are consumed. Therefore, if [Z] increases by 1.0 \, \text{mol L}^{-1} , then:
- Change in X = -\frac{1}{2} \times 1.0 = -0.5 \, \text{mol L}^{-1}
- Change in Y = -\frac{1}{2} \times 1.0 = -0.5 \, \text{mol L}^{-1}
### Step 4: Calculate the Equilibrium Concentrations
Now we can calculate the equilibrium concentrations of X and Y :
- [X] = 1.0 - 0.5 = 0.5 \, \text{mol L}^{-1}
- [Y] = 1.5 - 0.5 = 1.0 \, \text{mol L}^{-1}
### Step 5: Substitute into the Equilibrium Expression
Now we can substitute the equilibrium concentrations into the expression for K_c :
K_c = \frac{(1.0)^2}{(0.5)(1.0)} = \frac{1.0}{0.5} = 2.0
### Step 6: Relate to the Given Expression
We are given that K_c = \frac{x}{15} . Therefore, we can set up the equation:
2.0 = \frac{x}{15}
### Step 7: Solve for x
To find x , we multiply both sides by 15:
x = 2.0 \times 15 = 30
Thus, the value of x is:
\boxed{30}
