From the following Information, calculate the standard change of enthalpy for the combustion of coal. SHOW WORK AND BALANCED REACTION!!!
Compound DeltaHf (kj/mol)
Coal 0.0
Carbon dioxide -393.5
Water (l) -285.9
Oxygen 0.0
See your duplicate post below.
To calculate the standard change of enthalpy for the combustion of coal, we need to find the balanced reaction equation first. From the information given, the reactants are coal (C) and oxygen (O2), and the products are carbon dioxide (CO2) and water (H2O).
The balanced reaction equation for the combustion of coal can be written as:
C + O2 -> CO2 + H2O
To calculate the standard change of enthalpy (∆H°) for this reaction, we use the formula:
∆H° = Σ∆H°(products) - Σ∆H°(reactants)
Now let's calculate the standard change of enthalpy:
ΔH°(C) = 0.0 kJ/mol (given)
ΔH°(CO2) = -393.5 kJ/mol (given)
ΔH°(H2O) = -285.9 kJ/mol (given)
ΔH°(O2) = 0.0 kJ/mol (given)
Σ∆H°(products) = ΔH°(CO2) + ΔH°(H2O)
= -393.5 kJ/mol + (-285.9 kJ/mol)
= -679.4 kJ/mol
Σ∆H°(reactants) = ΔH°(C) + ΔH°(O2)
= 0.0 kJ/mol + 0.0 kJ/mol
= 0.0 kJ/mol
∆H° = Σ∆H°(products) - Σ∆H°(reactants)
= -679.4 kJ/mol - 0.0 kJ/mol
= -679.4 kJ/mol
Therefore, the standard change of enthalpy for the combustion of coal is -679.4 kJ/mol.