Question

Calculate the standard enthalpy of formation (ΔH°f) for butane (C4H10) given the following data:ΔH°f (kJ/mol)C(s)      0H2(g)    0C2H6(g) -84.7C3H8(g) -103.8Note: You may assume that butane undergoes complete combustion to form CO2(g) and H2O(l) and that the standard enthalpies of combustion for C4H10, CO2, and H2O are -2877.2, -393.5, and -285.8 kJ/mol, respectively.

Answer 1

To calculate the standard enthalpy of formation  Hf  for butane  C4H10 , we can use the following equation:Hf  C4H10  =  Hf  products  -  Hf  reactants The complete combustion of butane can be represented by the following balanced chemical equation:C4H10 g  + 13/2 O2 g   4 CO2 g  + 5 H2O l Now, we can calculate the standard enthalpy of formation for butane using the given data:Hf  C4H10  = [4  Hf  CO2  + 5  Hf  H2O ] - [Hf  C4H10  + 13/2  Hf  O2 ]Since the standard enthalpy of formation for elements in their standard state is zero, Hf  C s   = Hf  H2 g   = Hf  O2 g   = 0.We are given the standard enthalpies of combustion for C4H10, CO2, and H2O as -2877.2, -393.5, and -285.8 kJ/mol, respectively. We can use these values to find the standard enthalpies of formation for CO2 and H2O:Hf  CO2  = -393.5 kJ/molHf  H2O  = -285.8 kJ/molNow, we can plug these values into the equation:Hf  C4H10  = [4   -393.5  + 5   -285.8 ] - [Hf  C4H10 ]Hf  C4H10  = [-1574 +  -1429 ] - [Hf  C4H10 ]Hf  C4H10  = -3003 - Hf  C4H10 Now, we can use the given standard enthalpy of combustion for C4H10:-2877.2 = -3003 - Hf  C4H10 Hf  C4H10  = -3003 + 2877.2Hf  C4H10  = 125.8 kJ/molSo, the standard enthalpy of formation for butane  C4H10  is 125.8 kJ/mol.