The value of `K_(p)` for dissociation of `2HI hArr H_(2)+I_(2)` is `1.84xx10^(-2)`. If the equilibrium concentration of `H_(2)` is `0.4789` mol `L^(-1)`, calculate the concentration of `HI` at equilibrium.

Explanation

To solve the problem, we need to calculate the concentration of HI at equilibrium given the dissociation reaction and the equilibrium concentration of H2.

### Step-by-Step Solution:

1. **Write the balanced chemical equation:**
The dissociation of hydrogen iodide (HI) can be represented as:
\[ 2 \text{HI} \rightleftharpoons \text{H}_2 + \text{I}_2 \]

2. **Identify the equilibrium concentrations:**
We are given:
- The equilibrium concentration of H2, \([H_2] = 0.4789 \, \text{mol L}^{-1}\).
- Since the stoichiometry of the reaction shows that 1 mole of H2 is produced for every mole of I2 produced, we have:
\[ [I_2] = 0.4789 \, \text{mol L}^{-1} \]

3. **Let the equilibrium concentration of HI be \(x\):**
At equilibrium, we can denote the concentration of HI as \(x\). The equilibrium expression for the reaction can be written using the concentrations of the products and reactants:
\[ K_c = \frac{[H_2][I_2]}{[HI]^2} \]

4. **Substitute known values into the equilibrium expression:**
We know that \(K_c = K_p = 1.84 \times 10^{-2}\). Therefore, we can substitute the known concentrations into the equilibrium expression:
\[ 1.84 \times 10^{-2} = \frac{(0.4789)(0.4789)}{x^2} \]

5. **Simplify the equation:**
This simplifies to:
\[ 1.84 \times 10^{-2} = \frac{0.4789^2}{x^2} \]

6. **Calculate \(0.4789^2\):**
First, calculate \(0.4789^2\):
\[ 0.4789^2 = 0.2297 \, (\text{approximately}) \]

7. **Rearranging the equation to solve for \(x^2\):**
Rearranging gives:
\[ x^2 = \frac{0.2297}{1.84 \times 10^{-2}} \]

8. **Calculate \(x^2\):**
\[ x^2 = \frac{0.2297}{0.0184} \approx 12.48 \]

9. **Take the square root to find \(x\):**
\[ x = \sqrt{12.48} \approx 3.53 \, \text{mol L}^{-1} \]

10. **Conclusion:**
Therefore, the concentration of HI at equilibrium is approximately:
\[ [HI] \approx 3.53 \, \text{mol L}^{-1} \]