A 1.2 g sample of C-5 H-10 O is burned completely in excess oxygen to yield CO2(g) and H2O(l) at 25 C, releasing 43.52 kJ of energy. What is the enthalpy of formation of C-5 H-10 O?
delta H = DH.
DHrxn = (n*DHf products) - (n*DHf reactants).
DHrxn = 43,520 J.
n = the appropriate coefficients in the balanced equation for each compound you look up for DHf.
DHf = look up these values in your text or notes.
Post your work if you get stuck.
To find the enthalpy of formation of C-5 H-10 O, we need to use the given information about the combustion reaction and apply the concept of Hess's Law.
Hess's Law states that the change in enthalpy of a reaction is independent of the pathway taken as long as the initial and final conditions are the same. In this case, we can use the enthalpy change from the combustion reaction to calculate the enthalpy of formation.
The balanced equation for the combustion of C-5 H-10 O is:
C-5 H-10 O + 7.5 O2 -> 5 CO2 + 5 H2O
From the equation, we can see that the stoichiometric ratio between C-5 H-10 O and CO2 is 1:5. This means that for every mole of C-5 H-10 O burned, 5 moles of CO2 are produced.
First, we need to calculate the number of moles of C-5 H-10 O in the sample:
Molar mass of C-5 H-10 O = (5 * 12.01 g/mol) + (10 * 1.01 g/mol) + (16.00 g/mol) = 86.14 g/mol
Number of moles of C-5 H-10 O = mass / molar mass = 1.2 g / 86.14 g/mol = 0.0139 mol
Now, we can calculate the enthalpy change for the combustion reaction:
Enthalpy change = Energy released / moles of C-5 H-10 O burned
= -43.52 kJ / 0.0139 mol
= -3132.37 kJ/mol
The negative sign indicates that energy is being released, which is typical for an exothermic reaction.
Since the enthalpy change for the combustion reaction is the sum of the enthalpy changes for the formation of CO2 and H2O, we can express it with the enthalpy of formation of CO2 and H2O:
-3132.37 kJ/mol = 5 * enthalpy of formation of CO2 + 5 * enthalpy of formation of H2O
The enthalpy of formation of CO2 is known and can be looked up in a standard enthalpy of formation table. Its value is -393.5 kJ/mol.
By rearranging the equation, we can solve for the enthalpy of formation of C-5 H-10 O:
5 * enthalpy of formation of CO2 + 5 * (-393.5 kJ/mol) = -3132.37 kJ/mol
5 * enthalpy of formation of CO2 = -3132.37 kJ/mol + 5 * 393.5 kJ/mol
5 * enthalpy of formation of CO2 = -3132.37 kJ/mol + 1967.5 kJ/mol
5 * enthalpy of formation of CO2 = -1164.87 kJ/mol
enthalpy of formation of C-5 H-10 O = -1164.87 kJ/mol / 5 = -232.97 kJ/mol
Therefore, the enthalpy of formation of C-5 H-10 O is approximately -232.97 kJ/mol.