A 2.3 g sample of C5H10O is burned completely in excess oxygen to yield CO2(g) and H2O(l) at 25 C, releasing 83.42 kJ of energy. What is the enthalpy of formation of C5H10O?
What is the DHf for C5H10O?
5C + 5H2 + 1/2 O2 = C5H10O
The combustion reaction is what?
C5H10O + 7O2 ==> 5CO2 + 5H2O
DHf rxn = (5*DHf CO2 + 5*DHf H2O) - (DHf C5H10O)
Look up values for DHf for CO2 and H2O, plug into the equation above with DHf rxn and solve for DHf C5H10O
To find the enthalpy of formation of C5H10O, we need to use the equation:
∆H = Σ ∆Hf(products) - Σ ∆Hf(reactants)
Where ∆H is the enthalpy change of the reaction, ∆Hf(products) is the standard enthalpy of formation for the products, and ∆Hf(reactants) is the standard enthalpy of formation for the reactants.
In this case, the reactants are C5H10O and O2(g), and the products are CO2(g) and H2O(l). We need to consider the stoichiometric coefficients when determining the values.
First, we need to determine the number of moles of C5H10O used. We know that the sample weighs 2.3 g, and the molar mass of C5H10O is approximately 86.14 g/mol.
Number of moles of C5H10O = mass / molar mass
= 2.3 g / 86.14 g/mol
≈ 0.027 mol
Next, we need to convert the moles of C5H10O to moles of CO2 and H2O based on the balanced chemical equation. In this case, for every mole of C5H10O, we get 5 moles of CO2 and 5 moles of H2O.
Number of moles of CO2 = 0.027 mol × 5 = 0.135 mol
Number of moles of H2O = 0.027 mol × 5 = 0.135 mol
Now, we need to calculate the enthalpy change of the reaction using the given energy release of 83.42 kJ.
∆H = Σ ∆Hf(products) - Σ ∆Hf(reactants)
We know that the formation of CO2 and H2O are both exothermic processes, so their standard enthalpies of formation are negative.
∆H = ∆Hf(CO2) + ∆Hf(H2O) - ∆Hf(C5H10O)
Since we are looking for the enthalpy of formation of C5H10O, we can rearrange the equation as:
∆Hf(C5H10O) = ∆Hf(CO2) + ∆Hf(H2O) - ∆H
Now, let's substitute the known values into the equation:
∆Hf(C5H10O) = 0 kJ/mol + (-285.83 kJ/mol) - (-83.42 kJ / 0.027 mol)
= -1050.37 kJ/mol / 0.027 mol
≈ -38994.44 kJ/mol
Therefore, the enthalpy of formation of C5H10O is approximately -38994.44 kJ/mol.