Calculate the standard entropy change, Srxn, for the combustion of acetylene: 2

Question

Calculate the standard entropy change, Srxn, for the combustion of acetylene: 2 C2H2 (g) + 5 O2 (g) 4 CO2 (g) + 2 H2O (g), using standard entropy values below: (Report your answer in standard notation.)
So (J/molK) C2H2 (g) 200.9 O2 (g) 205.2 CO2 (g) 213.8 H2O (g) 188.8

Calculate the standard entropy change, Srxn, for the combustion of acetylene: 2 C2H2 (g) + 5 O2 (g) 4 CO2 (g) + 2 H2O (g), using standard entropy values below: (Report your answer in standard notation.)
So (J/molK) C2H2 (g) 200.9 O2 (g) 205.2 CO2 (g) 213.8 H2O (g) 188.8

Calculate the standard entropy change, Srxn, for the combustion of acetylene: 2 C2H2 (g) + 5 O2 (g) 4 CO2 (g) + 2 H2O (g), using standard entropy values below: (Report your answer in standard notation.)
So (J/molK) C2H2 (g) 200.9 O2 (g) 205.2 CO2 (g) 213.8 H2O (g) 188.8 Calculate the standard entropy change, Srxn, for the combustion of acetylene: 2 C2H2 (g) + 5 O2 (g) 4 CO2 (g) + 2 H2O (g), using standard entropy values below: (Report your answer in standard notation.)
So (J/molK) C2H2 (g) 200.9 O2 (g) 205.2 CO2 (g) 213.8 H2O (g) 188.8

Explanation / Answer

2C2H2(g) + 5O2(g) ---> 4CO2(g) + 2H2O(g)

DeltaSrxn = Sum of products - Sum.of reactants

Sum of products-

4*S°(CO2) + 2*S°(H2O) = 4*213.8 + 2*188.8 = 1232.8J/molK

Sum.of reactants-

2*S°(C2H2) + 5*S°(O2) = 2*200.9 + 5*205.2 = 1427.8J/molK

DeltaSrxn = 1232.8J/molK - 1427.8J/molK

DeltaSrxn. = -195J/molK

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