In a reaction between `H_(2)` and `I_(2)` at a certain temperature, the amounts of `H_(2), I_(2)` and HI at equilibrium were found to be `0.45` mol, `0.39` mol, and `3.0` mol respectively. Calculate the equilibrium constant for the reaction at the given temperature.

Explanation

To calculate the equilibrium constant \( K_c \) for the reaction between \( H_2 \) and \( I_2 \) forming \( HI \), we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation for the reaction is:
\[ H_2(g) + I_2(g) \rightleftharpoons 2 HI(g) \]

### Step 2: Write the expression for the equilibrium constant \( K_c \)
The expression for the equilibrium constant \( K_c \) is given by the formula:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]
where \([HI]\), \([H_2]\), and \([I_2]\) are the equilibrium concentrations of the respective species.

### Step 3: Substitute the equilibrium concentrations into the expression
From the problem, we have:
- \([H_2] = 0.45 \, \text{mol}\)
- \([I_2] = 0.39 \, \text{mol}\)
- \([HI] = 3.0 \, \text{mol}\)

Assuming the volume of the system is \( V \), the concentrations can be expressed as:
\[ [H_2] = \frac{0.45}{V}, \quad [I_2] = \frac{0.39}{V}, \quad [HI] = \frac{3.0}{V} \]

### Step 4: Substitute these values into the \( K_c \) expression
Substituting the concentrations into the \( K_c \) expression, we get:
\[ K_c = \frac{\left(\frac{3.0}{V}\right)^2}{\left(\frac{0.45}{V}\right)\left(\frac{0.39}{V}\right)} \]

### Step 5: Simplify the expression
This simplifies to:
\[ K_c = \frac{\frac{9.0}{V^2}}{\frac{0.45 \times 0.39}{V^2}} = \frac{9.0}{0.45 \times 0.39} \]

### Step 6: Calculate the value of \( K_c \)
Now we calculate \( 0.45 \times 0.39 \):
\[ 0.45 \times 0.39 = 0.1755 \]
Thus,
\[ K_c = \frac{9.0}{0.1755} \approx 51.28 \]

### Conclusion
The equilibrium constant \( K_c \) for the reaction at the given temperature is approximately:
\[ K_c \approx 51.28 \]