Using the Gibbs energy change, Δ G°=+63.3 kJ for the following reaction, Ag 2 CO 3(s) leftharpoons 2 Ag +(a q)+ CO 32-(a q) the Ks p of Ag 2 CO 3(s) in water at 25° C is (R=8.314 JK -1 mol -1)

Question

Using the Gibbs energy change, Δ G°=+63.3 kJ for the following reaction, Ag 2 CO 3(s) leftharpoons 2 Ag +(a q)+ CO 32-(a q) the Ks p of Ag 2 CO 3(s) in water at 25° C is (R=8.314 JK -1 mol -1) (A)  3.2 × 10-26  (B)  8.0 × 10-12  (C)  2.9 × 10-3  (D)  7.9 × 10-2

Tardigrade Admin · Accepted Answer

Δ G° is related to Ks p by the equation Δ G°=-2.303 R T log Ks p Given, Δ G°=+63.3 K J =63.3 × 103 J Thus, substitute Δ G°=63.3 × 103 J R=8.314 JK -1 mol -1 and T =298 K [25+273 K ] from the above equation to get 63.3 × 103=-2.303 × 8.314 × 298 log K sp ∴ log K sp =-11.09 ⇒ K sp = operatornameantilog(-11.09) K sp =8.0 × 10-12

Q. Using the Gibbs energy change, $Δ G^{\circ} = + 63.3 k J$ for the following reaction, $A g_{2} C O_{3} (s) ⇌ 2 A g^{+} (a q) + C O_{3}^{2 -} (a q)$ the $K_{s p}$ of $A g_{2} C O_{3} (s)$ in water at $2 5^{\circ} C$ is $(R = 8.314 J K^{- 1} m o l^{- 1})$

$3.2 \times 1 0^{- 26}$

$8.0 \times 1 0^{- 12}$

$2.9 \times 1 0^{- 3}$

$7.9 \times 1 0^{- 2}$

Q. Using the Gibbs energy change, ΔG∘=+63.3kJ for the following reaction, Ag2​CO3​(s)⇌2Ag+(aq)+CO32−​(aq) the Ksp​ of Ag2​CO3​(s) in water at 25∘C is (R=8.314JK−1mol−1)

3.2×10−26

8.0×10−12

2.9×10−3

7.9×10−2

Q. Using the Gibbs energy change, $Δ G^{\circ} = + 63.3 k J$ for the following reaction, $A g_{2} C O_{3} (s) ⇌ 2 A g^{+} (a q) + C O_{3}^{2 -} (a q)$ the $K_{s p}$ of $A g_{2} C O_{3} (s)$ in water at $2 5^{\circ} C$ is $(R = 8.314 J K^{- 1} m o l^{- 1})$

$3.2 \times 1 0^{- 26}$

$8.0 \times 1 0^{- 12}$

$2.9 \times 1 0^{- 3}$

$7.9 \times 1 0^{- 2}$