Question

Calorimetry and Hess’s Law Combined When 0.10 gram of C3H8 were burned in a calorimeter containing 75.00 grams of water. The temperature of the water rose from an initial temperature of 25.00 °C to a final temperature of 39.80 °C. Assume the heat absorbed by the calorimeter is negligible.

a. Write the equation for the combustion of C3H8 to form CO2 (g) and H2O (l).

b. Use the calorimetry data given to calculate deltaH for the combustion reaction in kJ/mole.

c. Use Hess' Law to Calculate ΔHf for C3H8 from the ΔHcomb you found from calorimetry. The ΔHf values for CO2(g) and H2O(l) are -393.5 kJ/mole and -285.8 kJ/mole respectively. (write all the ΔHf reactions and the ΔHcomb reaction. Keep in mind that you are calculating   ΔHf for C3H8)

Answer 1

The combustion of C3H8 is as follows:

C3H8 + O2 ==== CO2 + H2O

the balanced equation is

C₃H₈ + 5 O₂ ==== 3 CO₂ + 4 H₂O; you just need to balance the carbon first , second place the hydrogen and finally the oxygen.

now let´s calculate the Heat of reaction captured by the water:

heat capacity of water is 4.184 KJ / Kg K or 4.184 J / g * K

H = m Cp (T2 - T1)

H = 75 g * 4.184 * (39.8 - 25 ) = 4644.24 Joules

This is the heat gained by the water, the heat of reaction is the heat lost this is - 4644.24 Joules, statement says you have 0.1 grams of propane (mw = 44.1 g/gmol)

moles = mass / mw = 0.1 / 44.1 = 0.002267 moles, heat of reaction is

-4644.24 J / 0.002267 = -2, 048, 628.14 J / mole or -2 048.628 KJ / mol

To find the heat of reaction for C3H8, let´s look at the chemical equation:

C₃H₈ + 5 O₂ ==== 3 CO₂ + 4 H₂O

H rxn = 3 H _(CO2) + 4 H _(H2O) - H _(C3H8)

-2048.628 = 3 * (-393.5) + 4 * (-285.8) - H _C3H8

-2048.628 = -2323.7 - H C3H8

-2048.628 + 2323.7 = - H C3H8 = 275.07

H C3H8 (formation) = -275.07 KJ / mole