Using the data: C₂H₄(g), $\Delta$H°_f = +51.9 kJ mol^-1, S° = 219.8 J mol^-1 K^-1 CO₂(g), $\Delta$H°_f = -394 kJ mol^-1, S° = 213.6 J mol⁺¹ K^-1 H₂O(l), $\Delta$H°_f = -286.0 kJ mol^-1, S° = 69.96 J mol^-1 K^-1 O₂(g), $\Delta$H°_f = 0.00 kJ mol^-1, S° = 205 J mol^-1 K^-1 Calculate the maximum amount of work that can be obtained, at 25.0 °C, from the process:C₂H₄(g)+ 3 O₂(g)$\rarr$ 2 CO₂(g)+ 2 H₂O(l)

Question

Quizplus · Accepted Answer

The maximum amount of work that can be obtained from a process is given by the change in Gibbs free energy, $\Delta$G. $\Delta$G can be calculated using the equation:$\Delta$G = $\Delta$H - T$\Delta$Swhere $\Delta$H is the change in enthalpy, T is the temperature in Kelvin, and $\Delta$S is the change in entropy.For the given process, the change in enthalpy is:$\Delta$H = 2(-394 kJ mol^-1) + 2(-286.0 kJ mol^-1) - (+51.9 kJ mol^-1) = -1332 kJ mol^-1The change in entropy is:$\Delta$S = 2(213.6 J mol^-1 K^-1) + 2(69.96 J mol^-1 K^-1) - (219.8 J mol^-1 K^-1) = -13.8 J mol^-1 K^-1The temperature is 25.0 °C, which is 298.15 K.Therefore, the change in Gibbs free energy is:$\Delta$G = -1332 kJ mol^-1 - (298.15 K)(-13.8 J mol^-1 K^-1) = -1332 kJ mol^-1 + 3.9 kJ mol^-1 = -1328.1 kJ mol^-1The maximum amount of work that can be obtained from the process is therefore 1328.1 kJ mol^-1, which is closest to choice A) 1332 kJ mol^-1.

Using the Data: C2H4(g) Δ\DeltaΔ H°f = +51 Δ\DeltaΔ H°f = -394 KJ Mol-1, S° = 213

Using the Data:
C₂H₄(g) $\Delta$ H°_f = +51 $\Delta$ H°_f = -394 KJ Mol^-1, S° = 213