One mole of X and Y are allowed to react in a 2L container when equilinrium is reached the following reaction occurs `2X+YhArrZ`.
If the concentration of Z is 0.2M calculate the
equilibrium constant for this reaction

Options

A. 0.015
B. 2.22
C. 6. 70
D. 66. 7

Answer

To solve the problem, we need to calculate the equilibrium constant (Kc) for the reaction given the concentrations of the reactants and products at equilibrium. Let's break it down step by step.

### Step 1: Write the balanced chemical equation
The reaction given is:
\[ 2X + Y \rightleftharpoons Z \]

### Step 2: Determine initial concentrations
We start with 1 mole of \( X \) and 1 mole of \( Y \) in a 2L container. The initial concentrations can be calculated as follows:
- Concentration of \( X \) = \( \frac{1 \text{ mole}}{2 \text{ L}} = 0.5 \, M \)
- Concentration of \( Y \) = \( \frac{1 \text{ mole}}{2 \text{ L}} = 0.5 \, M \)
- Concentration of \( Z \) = \( 0 \, M \) (since it hasn't formed yet)

### Step 3: Set up the change in concentration at equilibrium
Let \( x \) be the change in concentration of \( Z \) at equilibrium. Given that the concentration of \( Z \) at equilibrium is \( 0.2 \, M \), we can write:
- Change in concentration of \( Z \) = \( +0.2 \, M \)
- Change in concentration of \( X \) = \( -2 \times 0.2 \, M = -0.4 \, M \) (because 2 moles of \( X \) are consumed for every mole of \( Z \))
- Change in concentration of \( Y \) = \( -0.2 \, M \) (because 1 mole of \( Y \) is consumed for every mole of \( Z \))

### Step 4: Calculate equilibrium concentrations
Now we can calculate the equilibrium concentrations:
- Concentration of \( X \) at equilibrium = \( 0.5 - 0.4 = 0.1 \, M \)
- Concentration of \( Y \) at equilibrium = \( 0.5 - 0.2 = 0.3 \, M \)
- Concentration of \( Z \) at equilibrium = \( 0.2 \, M \) (given)

### Step 5: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant expression for the reaction is:
\[ K_c = \frac{[Z]}{[X]^2[Y]} \]

### Step 6: Substitute the equilibrium concentrations into the expression
Now we substitute the equilibrium concentrations into the expression:
\[ K_c = \frac{0.2}{(0.1)^2 \times 0.3} \]

### Step 7: Calculate \( K_c \)
Calculating the denominator:
- \( (0.1)^2 = 0.01 \)
- \( 0.01 \times 0.3 = 0.003 \)

Now substituting back into the equation:
\[ K_c = \frac{0.2}{0.003} = 66.67 \]

### Final Answer
Thus, the equilibrium constant \( K_c \) for the reaction is approximately:
\[ K_c = 66.67 \]

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