Define heat of combustion - Leaving Cert Chemistry - Question Question (a) - 2003

To calculate the heat of combustion of propane (C₃H₈), we can use the heats of formation of the reactants and products. The reaction is as follows:

$C₃H₈ + 5O₂ → 3CO₂ + 4H₂O$

Using the heats of formation:

1. For the products:
   • Heat of formation of CO₂ = –394 kJ/mol (for 3 moles: $3 imes -394$)
   • Heat of formation of H₂O = –286 kJ/mol (for 4 moles: $4 imes -286$)

So, the total for products:

$ext{Total for products} = 3 imes (-394) + 4 imes (-286) = -1182 - 1144 = -2326 ext{ kJ/mol}$

2. For the reactants:
   • Heat of formation of C₃H₈ = -104 kJ/mol
   • Oxygen (O₂) is in its standard state and has a heat of formation of 0 kJ/mol.

Thus, the heat of formation of reactants (C₃H₈ + 5O₂):

$ext{Total for reactants} = -104 ext{ kJ/mol} + 0 = -104 ext{ kJ/mol}$

3. Now, we calculate the heat of combustion (ΔH) using:

$ext{ΔH} = ext{Total for products} - ext{Total for reactants}$

$ext{ΔH} = -2326 - (-104) = -2222 ext{ kJ/mol}$

Given that the heat of combustion of propane is –2222 kJ/mol, we can find the moles of propane needed to generate 500 kJ:

Using the formula:

ext{Number of moles} = rac{ ext{Energy required}}{ ext{Heat of combustion}}

Substituting the values:

ext{Number of moles} = rac{500}{2222} = 0.225 ext{ mol}

Now, to find the mass of propane, we use the molar mass of propane (C₃H₈):

• Molar mass of C₃H₈ = $3 imes 12.01 + 8 imes 1.008 = 44.1 ext{ g/mol}$$

Calculating the mass:

$ext{Mass} = ext{Number of moles} imes ext{Molar mass} = 0.225 imes 44.1 ext{ g} = 9.94 ext{ g}$

Rounding to the nearest gram, the mass of propane gas required is approximately 10 g.