Equilibrium constant for the following equilibrium is given at 0oC. Na2HPO4 .12H2O(s) leftharpoons Na2HPO4 . 7H2O(s) + 5H2O(g) Kp = 31.25 × 10-13 At equilibrium what will be partial pressure of water vapour:

Question

Equilibrium constant for the following equilibrium is given at 0oC. Na2HPO4 .12H2O(s) leftharpoons Na2HPO4 . 7H2O(s) + 5H2O(g) Kp = 31.25 × 10-13 At equilibrium what will be partial pressure of water vapour: (A) (1/5) × 10-3 atm (B) 0.5 × 10-3 atm (C) 5 × 10-2 atm (D) 5 × 10-3 atm

Tardigrade Admin · Accepted Answer

Na2HPO4 .12H2O(s) leftharpoons Na2HPO4 7H2O4 (s) + 5H2O (g) Kp = 31.25 × 10-13 KP = (PH2O)5 (PH2O)5 = 31.25 × 10-13 (PH2O) = (3125)1/5 × (10-15)1/5 (PH2O) = 5 × 10-3

Q. Equilibrium constant for the following equilibrium is given at $0^{o} C$ .
$N a_{2} H P O_{4} .12 H_{2} O (s) ⇌ N a_{2} H P O_{4} .7 H_{2} O (s) + 5 H_{2} O (g)$
$K_{p} = 31.25 \times 1 0^{- 13}$
At equilibrium what will be partial pressure of water vapour:

$\frac{1}{5} \times 1 0^{- 3}$ atm

$0.5 \times 1 0^{- 3}$ atm

$5 \times 1 0^{- 2}$ atm

$5 \times 1 0^{- 3}$ atm

$N a_{2} H P O_{4} .12 H_{2} O (s) ⇌ N a_{2} H P O_{4} 7 H_{2} O_{4} (s) + 5 H_{2} O (g)$
$K_{p} = 31.25 \times 1 0^{- 13}$
$K_{P} = (P_{H_{2} O})^{5}$
$(P_{H_{2} O})^{5} = 31.25 \times 10 - 13$
$(P_{H_{2} O}) = (3125)^{1/5} \times (1 0^{- 15})^{1/5}$
$(P_{H_{2} O}) = 5 \times 1 0^{- 3}$

Q. Equilibrium constant for the following equilibrium is given at 0oC. Na2​HPO4​.12H2​O(s)⇌Na2​HPO4​.7H2​O(s)+5H2​O(g) Kp​=31.25×10−13 At equilibrium what will be partial pressure of water vapour:

51​×10−3 atm

0.5×10−3 atm

5×10−2 atm

5×10−3 atm