A chemistry student needs to calculate the standard enthalpy change for the combustion reaction of methane gas (CH4) at 298 K and 1 atm pressure. The equation for the reaction is:CH4(g) + 2O2(g) --> CO2(g) + 2H2O(l)Given the following standard enthalpies of formation at 298 K:ΔHf(CO2) = -393.5 kJ/molΔHf(H2O) = -285.8 kJ/molΔHf(CH4) = -74.8 kJ/molCalculate ΔHrxn, the standard enthalpy change for the combustion reaction of methane gas.

Question

A chemistry student needs to calculate the standard enthalpy change for the combustion reaction of methane gas (CH4) at 298 K and 1 atm pressure. The equation for the reaction is:CH4(g) + 2O2(g) --> CO2(g) + 2H2O(l)Given the following standard enthalpies of formation at 298 K:ΔHf(CO2) = -393.5 kJ/molΔHf(H2O) = -285.8 kJ/molΔHf(CH4) = -74.8 kJ/m

Answer 1

To calculate the standard enthalpy change  Hrxn  for the combustion reaction of methane gas, 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.Hrxn = [Hf CO2  + 2 * Hf H2O ] - [Hf CH4  + 2 * Hf O2 ]Since the standard enthalpy of formation for an element in its standard state is zero, the Hf O2  term is 0.Hrxn = [ -393.5 kJ/mol  + 2 *  -285.8 kJ/mol ] - [ -74.8 kJ/mol  + 2 *  0 kJ/mol ]Hrxn =  -393.5 - 571.6 + 74.8  kJ/molHrxn = -890.3 kJ/molThe standard enthalpy change for the combustion reaction of methane gas  CH4  at 298 K and 1 atm pressure is -890.3 kJ/mol.