Using the following data, calculate equilibrium constant K for the reaction, AgCl (s) leftharpoons Ag (a q)++ Cl (a q)- Δ Gf( AgCl )°=-109.7 kJ Δ Gf( Ag +)°=77.1 kJ Δ Gf( Cl -)°=-131.2 kJ

Question

Using the following data, calculate equilibrium constant K for the reaction, AgCl (s) leftharpoons Ag (a q)++ Cl (a q)- Δ Gf( AgCl )°=-109.7 kJ Δ Gf( Ag +)°=77.1 kJ Δ Gf( Cl -)°=-131.2 kJ (A) 55.6 × 103 (B) 2.95 × 10-41 (C) 1.78 × 10-10 (D) 9.75

Tardigrade Admin · Accepted Answer

Δ G°= Sigma Δ Gf( text Product)°- Sigma Δ Gf text (Reactant)° =77.1+(-131.2)-(-109.7)=55.6 kJ =55.6 × 103 J We know that Δ G°=-2.303 R T log K log K=-(Δ G°/2.303 R T) =-(55.6 × 103/2.303 × 8.314 × 298)=-9.75 K=1.78 × 10-10

Q. Using the following data, calculate equilibrium constant $K$ for the reaction,
$A g C l_{(s)} ⇌ A g_{(a q)}^{+} + C l_{(a q)}^{-}$
$Δ G_{f (A g Cl)}^{\circ} = - 109.7 k J$
$Δ G_{f (A g^{+})}^{\circ} = 77.1 k J$
$Δ G_{f (C l^{-})}^{\circ} = - 131.2 k J$

$55.6 \times 1 0^{3}$

$2.95 \times 1 0^{- 41}$

$1.78 \times 1 0^{- 10}$

$9.75$

Q. Using the following data, calculate equilibrium constant K for the reaction, AgCl(s)​⇌Ag(aq)+​+Cl(aq)−​ ΔGf(AgCl)∘​=−109.7kJ ΔGf(Ag+)∘​=77.1kJ ΔGf(Cl−)∘​=−131.2kJ

55.6×103

2.95×10−41

1.78×10−10

9.75

ΔG∘=ΣΔGf( Product)∘​−ΣΔGf (Reactant)∘​ =77.1+(−131.2)−(−109.7)=55.6kJ=55.6×103J We know that ΔG∘=−2.303RTlogK logK=−2.303RTΔG∘​ =−2.303×8.314×29855.6×103​=−9.75 K=1.78×10−10

Q. Using the following data, calculate equilibrium constant $K$ for the reaction,
$A g C l_{(s)} ⇌ A g_{(a q)}^{+} + C l_{(a q)}^{-}$
$Δ G_{f (A g Cl)}^{\circ} = - 109.7 k J$
$Δ G_{f (A g^{+})}^{\circ} = 77.1 k J$
$Δ G_{f (C l^{-})}^{\circ} = - 131.2 k J$

$55.6 \times 1 0^{3}$

$2.95 \times 1 0^{- 41}$

$1.78 \times 1 0^{- 10}$

$9.75$