The equilibrium constant for the efflorescence reaction is at . What is the equilibrium partial pres — Chemical Equilibrium Chemistry Question
Question
The equilibrium constant $K_p$ for the efflorescence reaction $CaSO_4 \cdot 5H_2O(s) \rightleftharpoons CaSO_4 \cdot 3H_2O(s) + 2H_2O(g)$ is $2.25 \times 10^{-4}\text{ atm}^2$ at $298\text{ K}$. What is the equilibrium partial pressure of water vapor in atm multiplied by $100$?
Answer: 1.5
💡 Solution & Explanation
The expression for $K_p$ involves only the gaseous product, so $K_p = (P_{H_2O})^2$. Given $K_p = 2.25 \times 10^{-4}$, we take the square root to find $P_{H_2O} = \sqrt{2.25 \times 10^{-4}} = 1.5 \times 10^{-2}\text{ atm}$. The question asks for $100 \times P_{H_2O} = 100 \times 0.015 = 1.5$.
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