For `I_(2)(g) hArr 2I(g), K_(p)=1.79xx10^(-10)`. The partial pressure of `I_(2)=1.0` atm and `I=0.5xx10^(-6)` atm after `50 min`. Comment on the status of equilibrium process.

Explanation

To determine the status of the equilibrium process for the reaction \( I_2(g) \rightleftharpoons 2I(g) \) with \( K_p = 1.79 \times 10^{-10} \), we need to calculate the reaction quotient \( Q_p \) using the given partial pressures of the reactants and products.

### Step-by-Step Solution:

1. **Write the expression for the reaction quotient \( Q_p \)**:
\[ Q_p = \frac{(P_I)^2}{(P_{I_2})} \]
where \( P_I \) is the partial pressure of iodine (I) and \( P_{I_2} \) is the partial pressure of diatomic iodine (I2).

2. **Substitute the given values into the expression**:
- Given \( P_{I_2} = 1.0 \, \text{atm} \)
- Given \( P_I = 0.5 \times 10^{-6} \, \text{atm} \)

Now substituting these values into the \( Q_p \) expression:
\[ Q_p = \frac{(0.5 \times 10^{-6})^2}{1.0} \]

3. **Calculate \( Q_p \)**:
\[ Q_p = \frac{(0.25 \times 10^{-12})}{1.0} = 0.25 \times 10^{-12} = 2.5 \times 10^{-13} \]

4. **Compare \( Q_p \) with \( K_p \)**:
- Given \( K_p = 1.79 \times 10^{-10} \)
- Now we compare \( Q_p \) and \( K_p \):
\[ Q_p = 2.5 \times 10^{-13} \quad \text{and} \quad K_p = 1.79 \times 10^{-10} \]

5. **Determine the direction of the reaction**:
- Since \( Q_p < K_p \), this indicates that the reaction will shift to the right (towards the products) to reach equilibrium.

6. **Conclusion**:
- The system is not at equilibrium. The reaction will proceed in the forward direction to form more iodine (I) until equilibrium is reached.