Final answer:
The pressure of water vapor at equilibrium with a mixture of Na2SO4·10H2O and NaSO4 is approximately 1.36 x 10-3 atm.
Step-by-step explanation:
The equation for the dehydration of sodium sulfate 10-hydrate, Na2SO4·10H₂O, is:
Na2SO4·10H₂O(s) → Na2SO4 (s) + 10H₂O(g)
To find the pressure of water vapor at equilibrium, we need to know the equilibrium constant, Kp, which is given as 4.08 x 10-25 at 25 °C. Since Kp refers to the partial pressures of the products and reactants, we can use this value to calculate the pressure of water vapor at equilibrium.
Using the equation: Kp = (P(H2O))^10, where P(H2O) is the pressure of water vapor, we can rearrange the equation to solve for P(H2O):
P(H2O) = (Kp)^(1/10) = (4.08 x 10-25)^(1/10) = 1.36 x 10-3
Therefore, the pressure of water vapor at equilibrium with a mixture of Na2SO4 10H₂O and NaSO4 is approximately 1.36 x 10-3 atm.