Question

Consider the reaction 2C2H6(g) + 7O2(g)4CO2(g) + 6H2O(g) where S° rxn = 92.7 J/K Using standard thermodynamic data (in the Chemistry References), calculate the entropy change of the surroundings and the universe at 25°C. Ssurroundings = J K-1 mol-1 Suniverse = J K-1 mol-1

Answer 1

Answer

∆S°_surroundings = 9709 J K^-1 mol^-1

∆S°_universe = 9802 J K^-1 mol^-1

Explanation

2C2H6(g) + 7O2(g) -------> 4CO2(g) + 6H2O(g)

∆H°_rxn = n∆H°_f(products) - m∆H°_f(reactants)

= ( 4mol × ∆H°_f(CO2(g)) + 6mol × ∆H°_f(H2O(g))) - ( 2mol × ∆H°_f(C2H2(g)) + 7mol × ∆H°_f( O2(g)))

= ( 4mol × -393.5kJ/mol + 6mol × -248.1kJ/mol) - ( 2mol × -84kJ/mol)

= -3062.6kJ + 168kJ

= - 2894.6kJ

∆S°_surroundings = - ∆H/T

∆S°_surroundings = -( -2894.6kJ/298 15K)

∆S°_surrondings = 9709 J/K

∆S°_system = 92.7J/K

∆S°_universe = ∆S°_surrondings + ∆S_system

_{​​​​​​}∆S°_universe = 9709J/K + 92.7J/K

∆S°_universe = 9802J/K