Given the following thermochemical data, what is the enthalpy of formation ΔfHo for C2H5OH(l) C(s) + O2(g) → CO2(g) ΔHo = –393 kJ mol–1H2(g) + 1/2 O2(g) → H2O (g) ΔHo = -298 kJ mol–1C2H5OH +3O2 → 3H2O (g) + 2CO2(g) ΔHo =–1369 kJ mol–1 Use-full formula ΔrxnHo= ∑ ΔfHoproducts - ∑ ΔfHoreactants
Question
Given the following thermochemical data, what is the enthalpy of formation ΔfHo for C2H5OH(l) C(s) + O2(g) → CO2(g) ΔHo = –393 kJ mol–1H2(g) + 1/2 O2(g) → H2O (g) ΔHo = -298 kJ mol–1C2H5OH +3O2 → 3H2O (g) + 2CO2(g) ΔHo =–1369 kJ mol–1 Use-full formula ΔrxnHo= ∑ ΔfHoproducts - ∑ ΔfHoreactants
Solution
The enthalpy of formation ΔfHo for C2H5OH(l) can be calculated using the given thermochemical data and the formula ΔrxnHo= ∑ ΔfHoproducts - ∑ ΔfHoreactants.
First, let's write down the enthalpy changes for the given reactions:
- C(s) + O2(g) → CO2(g) ΔHo = –393 kJ mol–1
- H2(g) + 1/2 O2(g) → H2O (g) ΔHo = -298 kJ mol–1
- C2H5OH +3O2 → 3H2O (g) + 2CO2(g) ΔHo =–1369 kJ mol–1
The enthalpy of formation for C2H5OH(l) can be calculated by rearranging the formula to ΔfHo = ∑ ΔfHoproducts - ΔrxnHo.
The enthalpy of formation for the products (3H2O and 2CO2) can be calculated by multiplying the enthalpy of formation for each product by the number of moles of each product:
∑ ΔfHoproducts = (3 mol H2O * -298 kJ/mol) + (2 mol CO2 * -393 kJ/mol) = -894 kJ + -786 kJ = -1680 kJ
Substituting the values into the formula gives:
ΔfHo = -1680 kJ - (-1369 kJ) = -311 kJ/mol
Therefore, the enthalpy of formation ΔfHo for C2H5OH(l) is -311 kJ/mol.
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