Question

Calculate the standard enthalpy change for the combustion of ethane (C2H6) given the balanced chemical equation:C2H6(g) + 3.5 O2(g) → 2CO2(g) + 3H2O(l)Assume all reactants and products are in standard states and use the following standard enthalpy of formation values:ΔHf°(C2H6) = -84.7 kJ/molΔHf°(CO2) = -393.5 kJ/molΔHf°(H2O) = -285.8 kJ/mol

Answer 1

To calculate the standard enthalpy change for the combustion of ethane, we can use the following equation:H =  Hf products  -  Hf reactants For the products, we have 2 moles of CO2 and 3 moles of H2O: Hf products  =  2  -393.5 kJ/mol  +  3  -285.8 kJ/mol  = -787.0 kJ/mol - 857.4 kJ/mol = -1644.4 kJ/molFor the reactants, we have 1 mole of C2H6 and 3.5 moles of O2. Since the standard enthalpy of formation for O2 is zero  it is an element in its standard state , we only need to consider the enthalpy of formation for C2H6: Hf reactants  =  -84.7 kJ/mol  +  3.5  0 kJ/mol  = -84.7 kJ/molNow, we can calculate the standard enthalpy change:H =  Hf products  -  Hf reactants  = -1644.4 kJ/mol -  -84.7 kJ/mol  = -1559.7 kJ/molThe standard enthalpy change for the combustion of ethane is -1559.7 kJ/mol.