The standard Gibbs free energy change (Δ G°) at 25°C for the dissociation of N2O4(g) to NO2(g) is (given, equilibrium constant = 0.15, R = 8.314 J/K/mol)

Question

The standard Gibbs free energy change (Δ G°) at 25°C for the dissociation of N2O4(g) to NO2(g) is (given, equilibrium constant = 0.15, R = 8.314 J/K/mol) (A) 1.1 kJ (B) 4.7 kJ (C) 8.1 kJ (D) 38.2 kJ

Tardigrade Admin · Accepted Answer

Δ G°=-2.303 R T log K =-2.303 × 8.314 × 298 × log 0.15 =-2.303 × 8.314 × 298 ×(-0.82) =4678.7 J =4.67 kJ

Q. The standard Gibbs free energy change $(Δ G^{\circ})$ at 25 $^{\circ}$ C for the dissociation of $N_{2} O_{4} (g)$ to $N O_{2} (g)$ is (given, equilibrium constant = $0.15, R = 8.314 J / K / m o l)$

$1.1 k J$

$4.7 k J$

$8.1 k J$

$38.2 k J$

$Δ G^{\circ} = - 2.303 RT lo g K$
$= - 2.303 \times 8.314 \times 298 \times lo g 0.15$
$= - 2.303 \times 8.314 \times 298 \times (- 0.82)$
$= 4678.7 J = 4.67 k J$

Q. The standard Gibbs free energy change (ΔG∘) at 25∘C for the dissociation of N2​O4​(g) to NO2​(g) is (given, equilibrium constant = 0.15,R=8.314J/K/mol)

1.1kJ

4.7kJ

8.1kJ

38.2kJ