a)The combustion of propane, C3H8, occurs via the reaction C3H8(g)+5O2(g)3CO2(g)

Question

a)The combustion of propane, C3H8, occurs via the reaction C3H8(g)+5O2(g)3CO2(g)+4H2O(g) with heat of formation values given by the following table: Substance Hf (kJ/mol) C3H8 (g) -104.7 CO2(g) 393.5 H2O(g) 241.8 Calculate the enthalpy for the combustion of 1 mole of propane.

B)

For which of the following reactions is Hrxn equal to Hf of the product(s)?

You do not need to look up any values to answer this question.

Check all that apply.

Check all that apply.

2Li(s)+Cl2(g)2LiCl(s) Li(s)+12Cl2(g)LiCl(s) C(s,graphite)+O2(g)CO2(g) Li(s)+12Cl2(l)LiCl(s) BaCO3(s)BaO(s)+CO2(g) CO(g)+12O2(g)CO2(g)

Explanation / Answer

First write the equation of the combustion reaction:

C3H8(g) + 5O2(g) ---> 3CO2(g) + 4H2O(l) ...dH = -2220.1 kJ/mol

Delta H subscript (f) means "the enthalpy of formation" of the substance under the standard state conditions (25 C and 1 atm).

Enthalpies of formations of gaseous CO2 and liquid water are given. Heat of combustion (nethalpy of combustion) of propane is also known, then;
delta Hf for C3H8 can be calculated from the enthalpy of the reaction. By definition; the enthalpy of a reaction (delta Hr);
delta Hr = Total delta Hf (products) - Total delta Hf (reactants)

Note: The enthalpy of formation of the most stable form of the elements is zero.

dH(combustion) = [3dHf (CO2) + 4dHf (H2O)] - [5dHf (O2) + dHf (C3H8)]

-2220.1 = [3(-393.5) + 4(-285.3)] - [ 0 + dHf (C3H8)]

-2220.1 = [-1180.5) - 1141.2] - dHf (C3H8)
-2220.1 = - 2321.7 - dHf (C3H8)
dHf (C3H8) = 2220.1 - 2321.7 = - 101.6 kJ/mol

2)

Li(s)+12Cl2(g)LiCl(s) C(s,graphite)+O2(g)CO2(g)

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