CuSO4 . 5H2O(s) ⇌ CuSO4(s) + 5H2O(g) Kp = 10–10 (atm). 10–2 moles of CuSO4 . 5H2O(s) is taken in a 2.5L container at 27°C then at equilibrium [Take : R = \frac{1}{12} L atm mol–1 K

Engineering

Chemistry

Large or Small Equilibrium Constant Problems

Question

CuSO₄ . 5H₂O(s) ⇌ CuSO₄(s) + 5H₂O(g) K_p = 10^–10 (atm). 10^–2 moles of CuSO₄ . 5H₂O(s) is taken in a 2.5L container at 27°C then at equilibrium [Take : R = $\frac{1}{12}$ L atm mol^–1 K^–1]

Moles of CuSO₄ left in the container is 2 × 10^–4

Moles of CuSO₄ . 5H₂O left in the container is 9.8 × 10^–3

Moles of CuSO₄ . 5H₂O left in the container is 9 × 10^–3

Moles of CuSO₄ left in the container is 10^–3

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$10^{- 10} {atm}^{5} = P_{H_{2} O}^{5}$ ⇒ $P_{H_{2} O}^{} = 10^{- 2} atm .$

$n = \frac{PV}{RT} = \frac{10^{- 2} \times 2 .5}{\frac{1}{12} \times 300} = 10^{- 3}$

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