Question

Calculate the enthalpy change in kJ/mol for the combustion of butane (C4H10) given that the heat of formation of CO2(g) is -393.5 kJ/mol, the heat of formation of H2O(l) is -285.8 kJ/mol, and the heat of formation of C4H10(g) is -126 kJ/mol. The balanced equation for the combustion of butane is:2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l) + heat.

Answer 1

To calculate the enthalpy change for the combustion of butane, we will use the following equation:H_combustion =  H_f products  -  H_f reactants where H_combustion is the enthalpy change for the combustion of butane, H_f products  is the sum of the heats of formation of the products, and H_f reactants  is the sum of the heats of formation of the reactants.From the balanced equation, we have:2C4H10 g  + 13O2 g   8CO2 g  + 10H2O l First, let's find the sum of the heats of formation of the products:8 moles CO2 *  -393.5 kJ/mol  + 10 moles H2O *  -285.8 kJ/mol  = -3148 kJ - 2858 kJ = -6006 kJNow, let's find the sum of the heats of formation of the reactants:2 moles C4H10 *  -126 kJ/mol  + 13 moles O2 * 0 kJ/mol = -252 kJNote that the heat of formation of O2 is 0 kJ/mol because it is in its standard state.Now, we can find the enthalpy change for the combustion of butane:H_combustion = -6006 kJ -  -252 kJ  = -5754 kJSince the balanced equation involves 2 moles of butane, we need to divide the enthalpy change by 2 to find the enthalpy change per mole of butane:H_combustion per mole of butane = -5754 kJ / 2 = -2877 kJ/molTherefore, the enthalpy change for the combustion of butane is -2877 kJ/mol.