Question

The enthalpy of combustion of propane, graphite and dihydrogen at 298 K are –2220.0 kJ mol–1, –393.5 kJ mol–1 and –285.8 kJ mol–1, respectively.The magnitude of enthalpy of formation of propane (C3H8) is _____ kJ mol–1. (Nearest integer)

Answer 1

The enthalpy of the formation of propane can be calculated using Hess’s Law, which states that the enthalpy change of a chemical reaction is independent of the route by which the chemical reaction takes place. This is done through the manipulation of the enthalpy values provided. For propane, we know the balanced equation for its formation is: C(s) + H₂(g) $\rightarrow$ C₃H₈(l)
But we'll need 3 moles of carbon and 4 moles of hydrogen for this, so the equation should be: 3C(s) + 4H₂(g) $\rightarrow$ C₃H₈(l)
Hence, according to Hess’s Law, the enthalpy change of the formation of propane is:
ΔH = [ 1 x (–2220.0 kJ mol⁻¹) - 3 x (–393.5 kJ mol⁻¹) - 4 x (–285.8 kJ mol⁻¹)]
= –103.5 kJ mol⁻¹
$\Rightarrow$ round off to the nearest whole number
$\Rightarrow$ -104 kJ mol⁻¹
So, the correct answer is 104