Question

Calculate the standard enthalpy change for the reaction: CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g) Given the necessary bond enthalpies are: C-H = 411 kJ/molO=O = 495 kJ/molO-H = 463 kJ/molC=O = 799 kJ/mol.

Answer 1

To calculate the standard enthalpy change for the reaction, we need to find the difference between the total bond enthalpies of the reactants and the products.Reactants:1 CH4 molecule has 4 C-H bonds: 4 * 411 kJ/mol = 1644 kJ/mol2 O2 molecules have 2 O=O bonds: 2 * 495 kJ/mol = 990 kJ/molTotal bond enthalpy for reactants = 1644 kJ/mol + 990 kJ/mol = 2634 kJ/molProducts:1 CO2 molecule has 2 C=O bonds: 2 * 799 kJ/mol = 1598 kJ/mol2 H2O molecules have 4 O-H bonds: 4 * 463 kJ/mol = 1852 kJ/molTotal bond enthalpy for products = 1598 kJ/mol + 1852 kJ/mol = 3450 kJ/molStandard enthalpy change for the reaction = Total bond enthalpy of products - Total bond enthalpy of reactantsH = 3450 kJ/mol - 2634 kJ/mol = 816 kJ/molThe standard enthalpy change for the reaction is 816 kJ/mol.