What is the standard enthalpy change for the combustion of methane (CH4) if ΔHf

What is the standard enthalpy change for the combustion of methane (CH4) if ΔHf for CO2(g) is -393.5 kJ/mol and for H2O(l) is -285.8 kJ/mol?

What is the standard enthalpy change for the combustion of methane (CH4) if ΔHf for CO2(g) is -393.5 kJ/mol and for H2O(l) is -285.8 kJ/mol?

Step 1: Write the balanced chemical equation for the combustion of methane (CH4). The equation is: CH4 + 2 O2 -> CO2 + 2 H2O.
Step 2: Identify the products of the reaction, which are carbon dioxide (CO2) and water (H2O).
Step 3: Find the standard enthalpy of formation (ΔHf) values for the products: ΔHf for CO2(g) is -393.5 kJ/mol and for H2O(l) is -285.8 kJ/mol.
Step 4: Calculate the total enthalpy of the products. For 1 mole of CO2 and 2 moles of H2O, the calculation is: (1 * -393.5) + (2 * -285.8).
Step 5: Perform the multiplication: 2 * -285.8 = -571.6 kJ.
Step 6: Add the enthalpy values of the products: -393.5 + (-571.6) = -965.1 kJ.
Step 7: The standard enthalpy of formation for the reactants (CH4 and O2) is 0 for CH4 and 0 for O2, so we do not add anything for them.
Step 8: The final calculation for the standard enthalpy change (ΔH) is: ΔH = Total enthalpy of products - Total enthalpy of reactants = -965.1 - 0 = -965.1 kJ/mol.
Step 9: The standard enthalpy change for the combustion of methane (CH4) is -890.3 kJ/mol.

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