Consider the combustion of liquid methanol,
CH3OH (l)
CH3OH (l) +3/2O2 (g) = CO2 (g) + 2H2O(l) H = -726.5 kJ
What is for the reaction represented by this equation?
Express your answer using four significant figures
To find the enthalpy change (ΔH) for the given reaction, we need to subtract the sum of the enthalpies of the reactants from the sum of the enthalpies of the products.
Given:
CH3OH (l) + 3/2O2 (g) → CO2 (g) + 2H2O (l)
ΔH = -726.5 kJ
To determine the ΔH value for the reaction, we need to know the enthalpy values of the reactants and products. Looking at the balanced equation, we can assign the following enthalpy changes to each compound:
ΔH(CH3OH) + ΔH(O2) → ΔH(CO2) + ΔH(H2O)
Since we are given the ΔH value for the overall reaction (-726.5 kJ), we can assume that the enthalpies of formation for CO2 and H2O are known. Using standard enthalpy of formation values for these compounds, we can find their ΔH values.
The standard enthalpy of formation (ΔHf°) represents the enthalpy change when one mole of a compound is formed from its elements in their standard states (usually at 25°C and 1 atm pressure).
Using a reliable source or a thermodynamics handbook, we can find the following standard enthalpy of formation values:
ΔHf°(CO2) = -393.5 kJ/mol
ΔHf°(H2O) = -285.8 kJ/mol
By substituting these known values, we can write the equation as:
ΔH(CH3OH) + ΔH(O2) = ΔH(CO2) + ΔH(H2O)
ΔH(CH3OH) = ΔH(CO2) + ΔH(H2O) - ΔH(O2)
ΔH(CH3OH) = (-393.5 kJ/mol + 2 * (-285.8 kJ/mol)) - 0 kJ/mol
Evaluating the equation:
ΔH(CH3OH) = -726.5 kJ
Therefore, the enthalpy change for the combustion of liquid methanol is -726.5 kJ.