Calculate using the enthaly of formation,the enthalpy of combustion of methanol,CH3OH
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.
To calculate the enthalpy of combustion of methanol (CH3OH) using the enthalpy of formation, you need to know the enthalpy of formation of methanol (ΔHf°) and the balanced chemical equation for the combustion of methanol.
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.
To calculate the enthalpy of combustion of methanol (CH3OH) using the enthalpy of formation, we first need to determine the balanced chemical equation for the combustion reaction.
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.