For a reaction `2HI hArr H_(2)+I_(2)`, at equilibrium `7.8 g, 203.2 g`, and `1638.4 g` of `H_(2), I_(2)`, and HI, respectively were found. Calculate `K_(c)`.

Explanation

To calculate the equilibrium constant \( K_c \) for the reaction \( 2HI \rightleftharpoons H_2 + I_2 \), we will follow these steps:

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

### Step 2: Calculate the number of moles of each substance
We need to calculate the number of moles of \( H_2 \), \( I_2 \), and \( HI \) using the formula:
\[ \text{Number of moles} = \frac{\text{Given mass (g)}}{\text{Molar mass (g/mol)}} \]

- **For \( H_2 \)**:
- Given mass = 7.8 g
- Molar mass of \( H_2 = 2 \, \text{g/mol} \)
\[ \text{Moles of } H_2 = \frac{7.8 \, \text{g}}{2 \, \text{g/mol}} = 3.9 \, \text{mol} \]

- **For \( I_2 \)**:
- Given mass = 203.2 g
- Molar mass of \( I_2 = 254 \, \text{g/mol} \)
\[ \text{Moles of } I_2 = \frac{203.2 \, \text{g}}{254 \, \text{g/mol}} \approx 0.8 \, \text{mol} \]

- **For \( HI \)**:
- Given mass = 1638.4 g
- Molar mass of \( HI = 127 \, \text{g/mol} \)
\[ \text{Moles of } HI = \frac{1638.4 \, \text{g}}{127 \, \text{g/mol}} \approx 12.87 \, \text{mol} \]

### Step 3: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[H_2][I_2]}{[HI]^2} \]
Where \( [H_2] \), \( [I_2] \), and \( [HI] \) are the molar concentrations of the respective substances.

### Step 4: Assume a volume for the system
Let’s assume the volume of the container is \( V \) liters.

### Step 5: Calculate concentrations
The concentrations can be calculated as follows:
- Concentration of \( H_2 \):
\[ [H_2] = \frac{3.9 \, \text{mol}}{V} \, \text{mol/L} \]
- Concentration of \( I_2 \):
\[ [I_2] = \frac{0.8 \, \text{mol}}{V} \, \text{mol/L} \]
- Concentration of \( HI \):
\[ [HI] = \frac{12.87 \, \text{mol}}{V} \, \text{mol/L} \]

### Step 6: Substitute concentrations into the \( K_c \) expression
Substituting the concentrations into the \( K_c \) expression:
\[ K_c = \frac{\left(\frac{3.9}{V}\right) \left(\frac{0.8}{V}\right)}{\left(\frac{12.87}{V}\right)^2} \]

### Step 7: Simplify the expression
This simplifies to:
\[ K_c = \frac{3.9 \times 0.8}{12.87^2} \times \frac{V^2}{V^2} = \frac{3.12}{165.6969} \approx 0.0188 \]

### Final Answer
Thus, the equilibrium constant \( K_c \) is approximately:
\[ K_c \approx 0.019 \]