In this example problem, we will use the standard change in enthalpy for a reaction and the standard enthalpies of formation. Our problem reads: calculate the standard enthalpy of formation for methanol given the following reaction. The reaction is the combustion of methanol to produce carbon dioxide and water vapor. We are given the standard enthalpy of reaction. We are also given the standard enthalpies of formation for carbon dioxide gas and water vapor. To solve this problem, we can use the equation: The standard enthalpy of reaction = the summation of the moles of products times their standard enthalpies of formation minus the summation of the moles of reactants times their standard enthalpies of formation. For our reaction, the standard enthalpy of reaction equals two times the standard enthalpy of formation for carbon dioxide plus four times the standard enthalpy of formation for H2O gas, subtract the quantity 2 times the standard enthalpy of formation for methanol (which is what we're solving for) plus 3 times the standard enthalpy of formation for oxygen gas. We can plug the numbers into this equation. We already know that our standard enthalpy of reaction is -1277 kilojoules. We also know the standard enthalpy of formation for carbon dioxide and H2O gas. The standard enthalpy of formation for oxygen gas is zero because it's already in its standard state. Now, we can solve for the standard enthalpy of formation for methanol using algebra. When we rearrange for the standard enthalpy of formation for methanol and put the numbers into our calculator, we should get -238.6 kilojoules per mole.