Consider the combustion of natural gas. also called methane (CH.). CH_3(g) + 3O_
ID: 479180 • Letter: C
Question
Consider the combustion of natural gas. also called methane (CH.). CH_3(g) + 3O_2 (g) ACO_2(g) + 2H_2O(g) Using the thermodynamic data provided, answer the questions below. Be sure to use units in your work and final answers. 6. Calculate DM and DS. 7. Is this reaction always, never, or sometimes spontaneous? Do you expect it to be "more spontaneous" at lower or higher temperature? 8. Calculate for the combustion of methane at room temperature (a) and at the temperature that paper bums (b). (a) T = 22 degree C (b) T = 233 degree C 9. From this calculation, which of these reactions has a larger driving force to occur (which is "more spontaneous"). Is this what you expected? Give some discussion of why you expected this or why this may be surprising.Explanation / Answer
The balanced reaction for combustion of methane is CH4+ 2O2---àCO2+2H2O
Enthalpy change of reaction,deltaH = 1* enthalpy change ofCO2 +2* enthalpy change of water- {1* enthalpy change of CH4+2* enthalpy change of O2} = 1*(-213.7)+2*(-241.8)-{1*(-74.8)+2*0}=-622.5 Kj
Where 1,2, 1,2 are coefficients of CO2, H2O and CH4 and O2 respectively.
Similalrly , detlaS ( entropy change )= 213.7*1+2*188.72-{1*186.2+2*205}=-5.06 J/K
deltaG= deltaH- TdeltaS = -622.5*1000+298*5.06 =-620992 J= -620.992 Kj
when deltaG is –ve, the reaction is spontaneous.
Since entropy change is –ve, the reaction becomes less spontaneous at higher temperature since at higher temperature, the contribution of +ve term of TdeltaS become significant making it less spontaneous.
AT T=22 deg.c =22+273= 295, deltaG= -622.5 +295*5.06/1000 =-621 Kj
At T= 233 deg.c, T= 273+233= 506K, deltaG= -622.5+506*5.06/1000 =-619.94 Kj
Since deltaG is more –ve, at 22 deg.c, it is more spontaneous than at 233 deg.c