Calculating entropy changes for a chemical reaction atstandard conditions you ca

Question

Calculating entropy changes for a chemical reaction atstandard conditions you can find So for thereaction below. For the reaction whose reactants andproducts are:
reactants: CH4 (g) O2 (g) products: CO2 (g) H2O (g) a. calculate the standard entropy change(J/mol-K). The answer depends on how you balance the equation. For thisquestion balance the equation using the smallest ratio of WHOLEnumbers. b. Entropy and the second law of thermodynamics CalculateSouniv in J/mol-K for the aboveprocess. Calculating entropy changes for a chemical reaction atstandard conditions you can find So for thereaction below. For the reaction whose reactants andproducts are:
reactants: CH4 (g) O2 (g) products: CO2 (g) H2O (g) a. calculate the standard entropy change(J/mol-K). The answer depends on how you balance the equation. For thisquestion balance the equation using the smallest ratio of WHOLEnumbers. b. Entropy and the second law of thermodynamics CalculateSouniv in J/mol-K for the aboveprocess.

Explanation / Answer

CH4 (g) + 2O2(g) ..............> CO2 (g)+ 2H2O( g) S0rxn= {1molx(S0(CO2(g)) + 2mol x(S0 (H2O(g))} - {1mol x(S0(CH4(g)) +2molx(S0(O2(g))}              = {1molx(213.6J/mol.K) +2mol x(188.83J/mol.K)} -{ 1molx(186.3J/mol.K) +2molx (205.0J/mol.K)}               = {591.26J/K} -{ 596.3J/K}              =-5.04J/K ........................................ S0univ =S0rxn +S0surr To calculate : S0surr , we mustfind the H0rxn becauseS0surr = H0rxn /T            H0rxn ={ 1xH0 f (CO2(g)) +2xH0f (H2O(g)) } - {1xH0f (CH4(g)) + 2xH0f (O2(g)) }                = { 1mol x(-393.5kJ/mol) + 2mol x(-241.82kJ/mol) }- {1molx(-74.8kJ/mol) +2mol x(0kJ/mol)}                = -802.34kJ    Here , neglect the sign    S0surr =H0rxn /T                    = 802.3kJ / 298K                     =2.6924kJ/K                    = 2692.4J/K ...................................               S0univ = S0rxn +S0surr                             =  -5.04J/K + 2692.4J/K                              =2687.36J/K .............................. The mentioned units , by you were wrong . ........................................ S0univ =S0rxn +S0surr To calculate : S0surr , we mustfind the H0rxn becauseS0surr = H0rxn /T            H0rxn ={ 1xH0 f (CO2(g)) +2xH0f (H2O(g)) } - {1xH0f (CH4(g)) + 2xH0f (O2(g)) }                = { 1mol x(-393.5kJ/mol) + 2mol x(-241.82kJ/mol) }- {1molx(-74.8kJ/mol) +2mol x(0kJ/mol)}                = -802.34kJ    Here , neglect the sign    S0surr =H0rxn /T                    = 802.3kJ / 298K                     =2.6924kJ/K                    = 2692.4J/K ...................................               S0univ = S0rxn +S0surr                             =  -5.04J/K + 2692.4J/K                              =2687.36J/K               S0univ = S0rxn +S0surr                             =  -5.04J/K + 2692.4J/K                              =2687.36J/K .............................. The mentioned units , by you were wrong .

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