Calculate the heats of combustion for the following reactions from the standard enthalpies of formation listed in Appendix 2. (ΔH°f (HgO(s)) = -90.7 kJ/mol and ΔH°f (FeO(s)) = -272 kJ/mol.)
(a) 2 Fe(s) + O2(g) 2 FeO(s)
kJ
(b) 2 CH3OH(l) + 3 O2(g) 2 CO2(g) + 4 H2O(l)
Please answer the same question asked by Daniel for me again.
To calculate the heats of combustion for the given reactions, you need to use the standard enthalpies of formation (ΔH°f) of the substances involved in the reaction.
(a) 2 Fe(s) + O2(g) → 2 FeO(s)
To calculate the heat of combustion for this reaction, you need to subtract the sum of the standard enthalpies of formation of the reactants from the sum of the standard enthalpies of formation of the products.
First, calculate the enthalpy change for the reactant side:
ΔH°reactants = (2 * ΔH°f(Fe(s))) + ΔH°f(O2(g))
ΔH°reactants = (2 * 0 kJ/mol) + ΔH°f(O2(g)) (Since ΔH°f(Fe(s)) = 0 kJ/mol)
Next, calculate the enthalpy change for the product side:
ΔH°products = (2 * ΔH°f(FeO(s)))
Now, calculate the heat of combustion (ΔH°combustion):
ΔH°combustion = ΔH°products - ΔH°reactants
Substitute the given values:
ΔH°combustion = [(2 * ΔH°f(FeO(s)))] - [(2 * 0 kJ/mol) + ΔH°f(O2(g))]
ΔH°combustion = [(2 * -272 kJ/mol)] - [(2 * 0 kJ/mol) + ΔH°f(O2(g))]
(b) 2 CH3OH(l) + 3 O2(g) → 2 CO2(g) + 4 H2O(l)
Repeat the same steps for this reaction as well:
Calculate the enthalpy change for the reactant side:
ΔH°reactants = (2 * ΔH°f(CH3OH(l))) + (3 * ΔH°f(O2(g)))
Calculate the enthalpy change for the product side:
ΔH°products = (2 * ΔH°f(CO2(g))) + (4 * ΔH°f(H2O(l)))
Calculate the heat of combustion (ΔH°combustion):
ΔH°combustion = ΔH°products - ΔH°reactants
Substitute the given values and solve for ΔH°combustion.
Note: To get the values of ΔH°f for O2(g), CO2(g), H2O(l), and CH3OH(l), you need to refer to Appendix 2 as mentioned in the question.
To calculate the heats of combustion for the given reactions, we need to use the standard enthalpies of formation of the compounds involved. The heats of combustion can be calculated using the standard enthalpies of formation according to the equation:
ΔH = ΣnΔH°f(products) - ΣmΔH°f(reactants)
where ΔH is the heat of combustion, ΣnΔH°f(products) is the sum of the standard enthalpies of formation of the products multiplied by their stoichiometric coefficients, and ΣmΔH°f(reactants) is the sum of the standard enthalpies of formation of the reactants multiplied by their stoichiometric coefficients.
(a) 2 Fe(s) + O2(g) -> 2 FeO(s)
Given: ΔH°f (FeO(s)) = -272 kJ/mol
To calculate the heat of combustion for this reaction, we can use the equation mentioned above:
ΔH = (2 * ΔH°f(FeO(s))) - (0 * ΔH°f(O2(g)) + (0 * ΔH°f(Fe(s)))
= (2 * -272 kJ/mol) - (0 * 0 kJ/mol + 0 * 0 kJ/mol)
= -544 kJ/mol
Therefore, the heat of combustion for reaction (a) is -544 kJ/mol.
(b) 2 CH3OH(l) + 3 O2(g) -> 2 CO2(g) + 4 H2O(l)
Given: ΔH°f (CO2(g)) = -393.5 kJ/mol and ΔH°f (H2O(l)) = -285.8 kJ/mol
We can use the same equation to calculate the heat of combustion for this reaction:
ΔH = (2 * ΔH°f(CO2(g)) + 4 * ΔH°f(H2O(l))) - (2 * ΔH°f(CH3OH(l)) + 3 * ΔH°f(O2(g)))
= (2 * -393.5 kJ/mol + 4 * -285.8 kJ/mol) - (2 * 0 kJ/mol + 3 * 0 kJ/mol)
= -788 kJ/mol - 0 kJ/mol
= -788 kJ/mol
Therefore, the heat of combustion for reaction (b) is -788 kJ/mol.