2CH3OH + 3O2 ==> 2CO2 + 4H2O
DHrxn = (n*DHfproducts) - (n*DHfreactants).
Look up delta Hf for CO2 and H2O and CH3OH, substitute into the equation I gave you above and calculate delta H for the reaction.
DHrxn = (n*DHfproducts) - (n*DHfreactants).
Look up delta Hf for CO2 and H2O and CH3OH, substitute into the equation I gave you above and calculate delta H for the reaction.
The balanced chemical equation for the combustion of methanol is:
CH3OH(l) + 1.5O2(g) → CO2(g) + 2H2O(l)
Step 1: Find the enthalpy change (ΔH) of the combustion reaction using the enthalpy of formation:
ΔH = ΣnΔHf°(products) - ΣnΔHf°(reactants)
Where:
- ΣnΔHf°(products) is the sum of the enthalpies of formation of the products, multiplied by their stoichiometric coefficients.
- ΣnΔHf°(reactants) is the sum of the enthalpies of formation of the reactants, multiplied by their stoichiometric coefficients.
Step 2: Substitute the enthalpies of formation for the reactants and products into the equation:
ΔH = [ΔHf°(CO2) + 2ΔHf°(H2O)] - [ΔHf°(CH3OH) + 1.5ΔHf°(O2)]
Step 3: Look up the enthalpies of formation from a reliable source. The standard enthalpies of formation for methanol (CH3OH), carbon dioxide (CO2), water (H2O), and oxygen (O2) are as follows:
ΔHf°(CH3OH) = -201 kJ/mol
ΔHf°(CO2) = -394 kJ/mol
ΔHf°(H2O) = -286 kJ/mol
ΔHf°(O2) = 0 kJ/mol
Step 4: Plug in the values into the equation:
ΔH = [ (-394 kJ/mol) + 2(-286 kJ/mol) ] - [ (-201 kJ/mol) + 1.5(0 kJ/mol) ]
ΔH = [ -976 kJ/mol ] - [ -201 kJ/mol ]
ΔH = -775 kJ/mol
Therefore, the enthalpy of combustion of methanol (CH3OH) is -775 kJ/mol. The negative sign indicates that the reaction is exothermic, meaning it releases heat energy.
The general combustion equation for methanol can be written as:
CH3OH + O2 -> CO2 + H2O
Now, let's look up the enthalpy of formation values for each compound involved in the equation. The enthalpy of formation (ΔHf) represents the energy change when one mole of a compound is formed from its constituent elements in their standard states at a given temperature and pressure.
The enthalpy of formation values we need are:
ΔHf(CH3OH) = -238.6 kJ/mol (methanol)
ΔHf(CO2) = -393.5 kJ/mol (carbon dioxide)
ΔHf(H2O) = -285.8 kJ/mol (water)
Next, we can calculate the enthalpy of combustion (ΔHcomb) by using the difference in enthalpy of formation between the reactants and products:
ΔHcomb = Σ(ΔHf(products)) - Σ(ΔHf(reactants))
Substituting the values, we get:
ΔHcomb = [ΔHf(CO2) + ΔHf(H2O)] - ΔHf(CH3OH)
ΔHcomb = (-393.5 kJ/mol + -285.8 kJ/mol) - (-238.6 kJ/mol)
To find the actual value, we can perform the calculation:
ΔHcomb = -679.3 kJ/mol + 238.6 kJ/mol
ΔHcomb = -440.7 kJ/mol
Therefore, the enthalpy of combustion of methanol (CH3OH) is approximately -440.7 kJ/mol.