Calculate Δ G for the given reaction at 298 K under given conditions: N 2 O 4(g)=2 NO 2(g) The reaction is carried out at 298 K and 20 bar. 5 moles each of N 2 O 4 and NO 2 are taken initially. (Δ GN2 O4°=100 kJ / mol , Δ GN2°=50 kJ / mol )

Question

Calculate Δ G for the given reaction at 298 K under given conditions: N 2 O 4(g)=2 NO 2(g) The reaction is carried out at 298 K and 20 bar. 5 moles each of N 2 O 4 and NO 2 are taken initially. (Δ GN2 O4°=100 kJ / mol , Δ GN2°=50 kJ / mol ) (A) -50 kJ (B) zero (C) 150 kJ (D) 200 kJ

Tardigrade Admin · Accepted Answer

Δ G text Reaction = Sigma Δ Gf( text Product )°- Sigma Δ Gf( text Reactant )° =2 × 50-100=0

$- 50 k J$

zero

$150 k J$

$200 k J$

$Δ G_{Reaction} = ΣΔ G_{f (Product)}^{\circ} - ΣΔ G_{f (Reactant)}^{\circ}$
$= 2 \times 50 - 100 = 0$

Q. Calculate ΔG for the given reaction at 298K under given conditions : N2​O4(g)​=2NO2(g)​ The reaction is carried out at 298K and 20 bar. 5 moles each of N2​O4​ and NO2​ are taken initially. (ΔGN2​O4​∘​=100kJ/mol,ΔGN2​∘​=50kJ/mol)

−50kJ

zero

150kJ

200kJ

ΔGReaction ​=ΣΔGf( Product )∘​−ΣΔGf( Reactant )∘​ =2×50−100=0

$- 50 k J$

$150 k J$

$200 k J$

$Δ G_{Reaction} = ΣΔ G_{f (Product)}^{\circ} - ΣΔ G_{f (Reactant)}^{\circ}$
$= 2 \times 50 - 100 = 0$