For combustion of ethanol C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l), the amount of heat produced as measured in bomb calormeter, is 1364.47 kJ/mol at 25°C. Assuming all gases are behaving ideally, the enthalpy of combustion ΔCH, for ethanol will be : (R = 8.314 kJ/mol)
Question
For combustion of ethanol C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l), the amount of heat produced as measured in bomb calormeter, is 1364.47 kJ/mol at 25°C. Assuming all gases are behaving ideally, the enthalpy of combustion ΔCH, for ethanol will be : (R = 8.314 kJ/mol)
Solution
The enthalpy of combustion (ΔCH) can be calculated using the equation:
ΔCH = ΔU + ΔnRT
where:
- ΔU is the internal energy change (which is equal to the heat produced in a bomb calorimeter, -1364.47 kJ/mol in this case)
- Δn is the change in the number of moles of gas in the reaction
- R is the ideal gas constant (8.314 J/mol.K)
- T is the temperature in Kelvin (25°C = 298.15 K)
First, calculate Δn: Δn = (moles of gaseous products) - (moles of gaseous reactants) Δn = (2 moles of CO2) - (3 moles of O2) = -1
Then, substitute the values into the equation: ΔCH = -1364.47 kJ/mol + (-1 mol)(8.314 J/mol.K)(298.15 K)
Note: Convert R from J/mol.K to kJ/mol.K by dividing by 1000, so R = 0.008314 kJ/mol.K
ΔCH = -1364.47 kJ/mol + (-1 mol)(0.008314 kJ/mol.K)(298.15 K) ΔCH = -1364.47 kJ/mol - 2.48 kJ/mol ΔCH = -1366.95 kJ/mol
So, the enthalpy of combustion for ethanol is -1366.95 kJ/mol.
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