0 votes
30 views
in Chemical reactions by (138k points)
What is the enthalpy change for the combustion of 1 mole of methane gas (CH4) at standard conditions, given the following information: ΔHf(CH4) = -74.8 kJ/mol, ΔHf(H2O) = -285.8 kJ/mol and ΔHf(CO2) = -393.5 kJ/mol?

1 Answer

0 votes
by (138k points)
To calculate the enthalpy change for the combustion of 1 mole of methane gas  CH4  at standard conditions, we first need to write the balanced chemical equation for the combustion reaction:CH4 g  + 2 O2 g   CO2 g  + 2 H2O l Now, we can use Hess's Law to calculate the enthalpy change for this reaction. Hess's Law states that the enthalpy change for a reaction is equal to the sum of the enthalpy changes for the formation of the products minus the sum of the enthalpy changes for the formation of the reactants.H_combustion = [Hf CO2  + 2 * Hf H2O ] - [Hf CH4  + 2 * Hf O2 ]Since O2 is an element in its standard state, its enthalpy of formation is zero. Therefore, the equation becomes:H_combustion = [Hf CO2  + 2 * Hf H2O ] - [Hf CH4 ]Now, we can plug in the given values:H_combustion = [ -393.5 kJ/mol  + 2 *  -285.8 kJ/mol ] - [ -74.8 kJ/mol ]H_combustion =  -393.5 - 571.6 + 74.8  kJ/molH_combustion = -890.3 kJ/molSo, the enthalpy change for the combustion of 1 mole of methane gas  CH4  at standard conditions is -890.3 kJ/mol.

Related questions

Welcome to Sarvan Science Q&A, where you can ask questions and receive answers from other members of the community.
...