Chemistry Tutorial 5

1. Chemistry Tutorial 5 Question 12 Wang Jia 12S7F

2. 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:

3. (a) Calculate the standard entropy change of formation, ∆SfΘ, in J mol-1, of CO(g) and of CO2(g). ∆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.)

4. (a) Calculate the standard entropy change of formation, ∆SfΘ, in J mol-1, of CO(g) and of CO2(g). ∆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.)

5. (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.

6. 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

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

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

9. ∆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.

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

11. Thank you !