To determine the final pressure of water vapor in the given system, we need to analyze the reaction and the conditions provided. We have two solid phases of A, A.5H2O(s) and A.3H2O(s), and they are in equilibrium with water vapor. The reaction can be represented as:
A.5H2O(s) ⇌ A.3H2O(s) + 2H2O(g)
Given that we have 1 millimole of each solid and the system is at 80% relative humidity, we can start by calculating the saturation pressure of water vapor at 300 K, which is given as 85 mm Hg. The relative humidity indicates how much water vapor is present compared to the maximum it can hold at that temperature.
Understanding the Initial Conditions
At 80% relative humidity, the partial pressure of water vapor in the moist air can be calculated as:
Partial Pressure of H2O = Relative Humidity × Saturation Pressure
Substituting the values:
Partial Pressure of H2O = 0.80 × 85 mm Hg = 68 mm Hg
Equilibrium Considerations
Next, we need to consider the equilibrium established by the reaction. Initially, we have:
- 1 millimole of A.5H2O(s)
- 1 millimole of A.3H2O(s)
- Water vapor at a partial pressure of 68 mm Hg
As the reaction proceeds, A.5H2O(s) will convert to A.3H2O(s) and water vapor. The reaction produces 2 moles of water vapor for every mole of A.5H2O(s) that reacts. Therefore, if we let x be the amount of A.5H2O(s) that reacts, the changes in moles will be:
- A.5H2O(s): 1 - x
- A.3H2O(s): 1 + x
- H2O(g): 68 mm Hg + 2x (in mm Hg)
Applying the Equilibrium Constant
The equilibrium constant (K) for this reaction can be expressed in terms of the partial pressures of the gaseous products and the solids, which do not appear in the expression:
K = (P(H2O)^2) / 1
At equilibrium, we can express K in terms of the initial conditions and the changes due to the reaction:
K = (68 + 2x)^2
However, we need to find the value of x that satisfies the equilibrium condition. Since we don't have a specific value for K, we can assume that the reaction will proceed until the partial pressure of water vapor reaches the saturation pressure of 85 mm Hg.
Final Calculation
Setting the final pressure of water vapor equal to the saturation pressure:
68 + 2x = 85
Solving for x:
2x = 85 - 68
2x = 17
x = 8.5 mm Hg
Now, substituting back to find the final pressure of water vapor:
Final Pressure of H2O = 68 mm Hg + 2(8.5 mm Hg) = 68 mm Hg + 17 mm Hg = 85 mm Hg
Conclusion
The final pressure of water vapor in the system, after the reaction reaches equilibrium, is 85 mm Hg. This indicates that the system has reached saturation with respect to water vapor at the given temperature and conditions.
