At what temperature will the equilibrium constant for the formation of NOCl ( g ) be K eq = K p =1.0 × 103 2 NO ( g )+ Cl 2( g ) longrightarrow 2 NOCl ( g ) Δ H °=-77.1 kJ mol -1 Δ S°=-121.2 J mol -1 K -1

Question

At what temperature will the equilibrium constant for the formation of NOCl ( g ) be K eq = K p =1.0 × 103 2 NO ( g )+ Cl 2( g ) longrightarrow 2 NOCl ( g ) Δ H °=-77.1 kJ mol -1 Δ S°=-121.2 J mol -1 K -1 (A) 432 K (B) 298 K (C) 412 K (D) 398 K

Tardigrade Admin · Accepted Answer

Δ G°=Δ H°-T Δ S°-2.303 R T log K text eq =Δ H°-T Δ S° T[Δ S°-2.303 R log K text eq ]=Δ H° ∴ T=(Δ H°/Δ S°-2.303 R log Ke q) =(-77.1 kJ/-121.2×10-3kJ - 2.3030×8.314×10-3 kJ mol-1 K-1 log 1.0×103) =432 K

Q. At what temperature will the equilibrium constant for the formation of NOCl(g) be Keq​=Kp​=1.0×103 2NO(g)+Cl2​(g)⟶2NOCl(g) ΔH∘=−77.1kJmol−1 ΔS∘=−121.2Jmol−1K−1

432 K

298 K

412 K

398 K

ΔG∘=ΔH∘−TΔS∘−2.303RTlogKeq ​=ΔH∘−TΔS∘ T[ΔS∘−2.303RlogKeq ​]=ΔH∘ ∴T=ΔS∘−2.303RlogKeq​ΔH∘​ =−121.2×10−3kJ−2.3030×8.314×10−3kJmol−1K−1log1.0×103−77.1kJ​=432K

Q. At what temperature will the equilibrium constant for the formation of $NOCl (g)$ be $K_{e q} = K_{p} = 1.0 \times 1 0^{3}$
$2 NO (g) + C l_{2} (g) ⟶ 2 NOCl (g)$
$Δ H^{\circ} = - 77.1 k J m o l^{- 1}$
$Δ S^{\circ} = - 121.2 J m o l^{- 1} K^{- 1}$