Question

Calculate the enthalpy change for the combustion of methane (CH4) at constant pressure, given that the heat of formation for methane is -74.9 kJ/mol and the balanced chemical equation is: CH4 (g) + 2O2 (g) -> CO2 (g) + 2H2O (l) The heat of formation for CO2 is -393.5 kJ/mol and the heat of formation for H2O is -285.8 kJ/mol.

Answer 1

To calculate the enthalpy change for the combustion of methane, we can use Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the products minus the sum of the enthalpy changes for the reactants.H_combustion =  Hf_products -  Hf_reactantsFor the given reaction:CH4  g  + 2O2  g  -> CO2  g  + 2H2O  l The enthalpy change for the products is:Hf_products =  1 mol CO2  -393.5 kJ/mol  +  2 mol H2O  -285.8 kJ/mol  = -393.5 kJ - 571.6 kJ = -965.1 kJThe enthalpy change for the reactants is:Hf_reactants =  1 mol CH4  -74.9 kJ/mol  +  2 mol O2  0 kJ/mol  = -74.9 kJNow, we can calculate the enthalpy change for the combustion of methane:H_combustion = Hf_products - Hf_reactants = -965.1 kJ -  -74.9 kJ  = -890.2 kJTherefore, the enthalpy change for the combustion of methane at constant pressure is -890.2 kJ/mol.