Chemistry Tutorial 5 Question 12 Wang Jia 12S7F

The standard enthalpy change of formation, ∆HfΘ, and the standard free energy change of formation, ∆GfΘ, of CO(g) and of CO2(g) are as follows: CO(g) CO2(g) ∆HfΘ /kJ mol -1 -110.5 -393.5 ∆GfΘ /kJ mol -1 -137.2 -394.4

(a) Calculate the standard entropy change of formation, ∆SfΘ, in J mol-1, of CO(g) and of CO2(g). ∆HfΘ /kJ mol -1 -110.5 -393.5 ∆GfΘ /kJ mol -1 -137.2 -394.4 ∆GfΘ = ∆HfΘ -T ∆SfΘ ∆GfΘ [CO(g)] = ∆HfΘ [CO(g)] -T ∆SfΘ [CO(g)] -137.2= -110.5 –(298) ∆SfΘ [CO(g)] ∆SfΘ [CO(g)] = 0.08960kJ mol-1 = +89.6 J mol-1 (3s.f.)

(a) Calculate the standard entropy change of formation, ∆SfΘ, in J mol-1, of CO(g) and of CO2(g). ∆HfΘ /kJ mol -1 -110.5 -393.5 ∆GfΘ /kJ mol -1 -137.2 -394.4 ∆GfΘ = ∆HfΘ -T ∆SfΘ ∆GfΘ [CO2(g)] = ∆HfΘ [CO2(g)] -T ∆SfΘ [CO2(g)] -394.4= -393.5 –(298) ∆SfΘ [CO2(g)] ∆SfΘ [CO2(g)] = +3.02J mol-1 (3s.f.)

(b) Using the ∆HfΘ values given and your answers in (a), show that the reaction: C(s) + CO2 2CO(g) is not feasible at 298Kand calculate the minimum temperature at which reaction becomes feasible. CO(g) CO2(g) ∆HfΘ /kJ mol -1 -110.5 -393.5 ∆GfΘ /kJ mol -1 -137.2 -394.4 ∆SfΘ /J mol -1 +89.60 +3.020

To show that the reaction between C(s) and CO2(g) is not feasible at 298K Calculate the standard Gibbs Free Energy of this reaction, ∆GΘ Positive value  not feasible Values that we need for the calculation: ∆HrΘ & ∆SrΘ

C(s) + CO2(g) 2CO(g) 2C(s) + O2(g) By Hess’s Law, ∆HrΘ C(s) + CO2(g) 2CO(g) 2C(s) + O2(g) ∆HfΘ [CO2(g)] =-393.5kJ mol-1 2∆HfΘ [CO(g)]=2( -110.5)kJ mol-1 By Hess’s Law, ∆HrΘ = -(-393.5) + 2(-110.5) =+172.5 kJ mol-1

C(s) + CO2(g) 2CO(g) 2C(s) + O2(g) By Hess’s Law, ∆SrΘ 2∆SfΘ[CO(g)]=2(+89.60 J mol-1 ∆SfΘ [CO2(g)] = +3.020 J mol-1 C(s) + CO2(g) 2CO(g) 2C(s) + O2(g) By Hess’s Law, ∆SrΘ = -3.020+ 2(89.60) =+176.2 J mol-1

∆GΘ = ∆HΘ -T ∆SΘ ∆GΘ = +172.5- 298(176.2/1000) = +120 kJ mol-1 Hence, the reaction is not feasible at 298K as the Gibbs Free energy of the reaction is positive at 298K.

For the reaction to be feasible, ∆G < 0 ∆H -T ∆S < 0 +172.5 – T (176.2/1000) < 0 T > 979 K

Thank you !