To calculate the equilibrium constant \( K_c \) for the reaction:
\[
A_2(g) + 2B_2(g) \rightleftharpoons 2AB_2(g) + \text{Heat}
\]
given the equilibrium concentrations (in moles) of \( A_2 \), \( B_2 \), and \( AB_2 \), we will follow these steps:
### Step 1: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) is given by the formula:
\[
K_c = \frac{[AB_2]^2}{[A_2][B_2]^2}
\]
where \([AB_2]\), \([A_2]\), and \([B_2]\) are the molar concentrations of the respective gases at equilibrium.
### Step 2: Calculate the total number of moles
From the problem, we have:
- Moles of \( A_2 = 5 \)
- Moles of \( B_2 = 3 \)
- Moles of \( AB_2 = 2 \)
Total moles at equilibrium:
\[
\text{Total moles} = 5 + 3 + 2 = 10 \text{ moles}
\]
### Step 3: Calculate the volume of the system using the ideal gas law
Using the ideal gas law \( PV = nRT \), we can rearrange it to find the volume \( V \):
\[
V = \frac{nRT}{P}
\]
Where:
- \( n = 10 \) moles (total moles)
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- \( T = 300 \, \text{K} \)
- \( P = 8.21 \, \text{atm} \)
Substituting the values:
\[
V = \frac{10 \times 0.0821 \times 300}{8.21}
\]
Calculating this gives:
\[
V \approx \frac{246.3}{8.21} \approx 30.0 \, \text{L}
\]
### Step 4: Calculate the concentrations
Now we can calculate the concentrations:
\[
[A_2] = \frac{5 \, \text{moles}}{30.0 \, \text{L}} = 0.167 \, \text{M}
\]
\[
[B_2] = \frac{3 \, \text{moles}}{30.0 \, \text{L}} = 0.100 \, \text{M}
\]
\[
[AB_2] = \frac{2 \, \text{moles}}{30.0 \, \text{L}} = 0.067 \, \text{M}
\]
### Step 5: Substitute the concentrations into the \( K_c \) expression
Now substitute the concentrations back into the \( K_c \) expression:
\[
K_c = \frac{(0.067)^2}{(0.167)(0.100)^2}
\]
Calculating this gives:
\[
K_c = \frac{0.004489}{(0.167)(0.010)} = \frac{0.004489}{0.00167} \approx 2.68
\]
### Step 6: Final answer
Thus, the equilibrium constant \( K_c \) for the reaction is approximately:
\[
K_c \approx 2.66
\]
