The maximum work ( in kJ mol -1 ) that can be derived from complete combustion of 1 mole of CO at 298 K and 1 atm is. [ Standard enthalpy of combustion of CO =-283.0 kJ mol -1; standard molar entropies at 298 K: S O 2=205.1 J mol -1, S CO =197.7 J mol -1 S CO 2=213.7 J mol-1]

Question

The maximum work ( in kJ mol -1 ) that can be derived from complete combustion of 1 mole of CO at 298 K and 1 atm is. [ Standard enthalpy of combustion of CO =-283.0 kJ mol -1; standard molar entropies at 298 K: S O 2=205.1 J mol -1, S CO =197.7 J mol -1 S CO 2=213.7 J mol-1] (A) 257 (B) 227 (C) 57 (D) 127

Tardigrade Admin · Accepted Answer

For the reaction, CO (g)+(1/2) O 2(g) longrightarrow CO 2(g)Δ HC=-283.5 Δ S = Sigma SP- Sigma SR =213.7-(197.7+(1/2) × 205.1) =-86.5 J mol -1 Δ G =Δ H-T Δ S =-283-298 ×-86.5 × 10-3 =-257.0 kJ mol -1 w max =-Δ G=-(-2570.0)=257 kJ mol -1

Q. The maximum work ( in kJmol−1 ) that can be derived from complete combustion of 1 mole of CO at 298K and 1 atm is. [ Standard enthalpy of combustion of CO=−283.0kJmol−1; standard molar entropies at 298K:SO2​​=205.1Jmol−1,SCO​=197.7Jmol−1SCO2​​=213.7Jmol−1]

257

227

57

127

For the reaction, CO(g)+21​O2​(g)⟶CO2​(g)ΔHC​=−283.5 ΔS=ΣSP​−ΣSR​ =213.7−(197.7+21​×205.1) =−86.5Jmol−1 ΔG=ΔH−TΔS =−283−298×−86.5×10−3 =−257.0kJmol−1 wmax​=−ΔG=−(−2570.0)=257kJmol−1