When 3.00 mole of A and 1.00 mole of B are mixed in a 1,00 litre vessel , the following reaction takes place
`A(g) +B(g) hArr 2C(g)`
the equilibrium mixture contains 0.5 mole of C. What is the value of equilibrium constant for the reaction ?

Options

A. 0.12
B. 6
C. 1.5
D. 3

Answer

To find the equilibrium constant (K) for the reaction \( A(g) + B(g) \rightleftharpoons 2C(g) \), we will follow these steps:

### Step 1: Write the balanced equation
The balanced equation for the reaction is:
\[ A(g) + B(g) \rightleftharpoons 2C(g) \]

### Step 2: Set up the initial concentrations
Initially, we have:
- Moles of A = 3.00
- Moles of B = 1.00
- Moles of C = 0.00

Since the volume of the vessel is 1.00 L, the initial concentrations are:
- \([A]_{initial} = 3.00 \, \text{mol/L}\)
- \([B]_{initial} = 1.00 \, \text{mol/L}\)
- \([C]_{initial} = 0.00 \, \text{mol/L}\)

### Step 3: Define the change in concentration
Let \( x \) be the amount of A and B that reacts. According to the stoichiometry of the reaction:
- For every 1 mole of A that reacts, 1 mole of B reacts, and 2 moles of C are produced.

At equilibrium, we have:
- Moles of A = \( 3.00 - x \)
- Moles of B = \( 1.00 - x \)
- Moles of C = \( 2x \)

### Step 4: Use the information given
We know that at equilibrium, the concentration of C is 0.5 moles. Therefore:
\[ 2x = 0.5 \]
From this, we can solve for \( x \):
\[ x = \frac{0.5}{2} = 0.25 \]

### Step 5: Calculate equilibrium concentrations
Now we can find the equilibrium concentrations:
- \([A]_{eq} = 3.00 - 0.25 = 2.75 \, \text{mol/L}\)
- \([B]_{eq} = 1.00 - 0.25 = 0.75 \, \text{mol/L}\)
- \([C]_{eq} = 0.5 \, \text{mol/L}\)

### Step 6: Write the expression for the equilibrium constant
The equilibrium constant \( K \) is given by the formula:
\[ K = \frac{[C]^2}{[A][B]} \]

### Step 7: Substitute the equilibrium concentrations into the expression
Substituting the equilibrium concentrations into the expression for \( K \):
\[ K = \frac{(0.5)^2}{(2.75)(0.75)} \]

### Step 8: Calculate the value of K
Calculating the numerator:
\[ (0.5)^2 = 0.25 \]

Calculating the denominator:
\[ (2.75)(0.75) = 2.0625 \]

Now substituting these values into the equation for \( K \):
\[ K = \frac{0.25}{2.0625} \approx 0.1219 \]

### Step 9: Round the value
Rounding to two decimal places, we find:
\[ K \approx 0.12 \]

Thus, the equilibrium constant for the reaction is approximately \( 0.12 \).

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