At `700 K` equilibrium constant for the reaction, `H_(2(g))+I_(2(g))hArr2HI_((g))`
is `54.8`. If `0.5 mol litre^(-1)` of `HI_((g))` is present at equilibrium at `700 K`, what are the concentrations of `H_(2(g))` and `I_(2(g))`, assuming that we initially started with `HI_((g))` and allowed it to reach equilibrium at `700 K`.

Explanation

To solve the problem, we will follow these steps:

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

### Step 2: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant expression for the reaction is given by:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]

### Step 3: Substitute known values into the \( K_c \) expression
We know that at equilibrium:
- \( [HI] = 0.5 \, \text{mol/L} \)
- \( K_c = 54.8 \)

Substituting these values into the expression gives:
\[ 54.8 = \frac{(0.5)^2}{[H_2][I_2]} \]

### Step 4: Recognize the relationship between \( [H_2] \) and \( [I_2] \)
Since the stoichiometry of the reaction shows that 1 mole of \( H_2 \) reacts with 1 mole of \( I_2 \) to produce 2 moles of \( HI \), we can assume:
\[ [H_2] = [I_2] = x \]

### Step 5: Rewrite the equilibrium expression using \( x \)
Now we can rewrite the equilibrium expression as:
\[ 54.8 = \frac{(0.5)^2}{x^2} \]

### Step 6: Solve for \( x^2 \)
Rearranging the equation gives:
\[ x^2 = \frac{(0.5)^2}{54.8} \]

Calculating \( (0.5)^2 \):
\[ (0.5)^2 = 0.25 \]

Now substituting this value:
\[ x^2 = \frac{0.25}{54.8} \]

### Step 7: Calculate \( x^2 \)
Calculating \( x^2 \):
\[ x^2 = 0.004564 \]

### Step 8: Find \( x \)
Taking the square root of both sides gives:
\[ x = \sqrt{0.004564} \]
\[ x \approx 0.0677 \, \text{mol/L} \]

### Step 9: Conclusion
Thus, the concentrations of \( H_2 \) and \( I_2 \) at equilibrium are:
\[ [H_2] = [I_2] \approx 0.068 \, \text{mol/L} \]

### Summary of Results
- \( [H_2] \approx 0.068 \, \text{mol/L} \)
- \( [I_2] \approx 0.068 \, \text{mol/L} \)