Question

2-propanol has the formula C₃H₈O and its combustion can be written as:

C₃H₈O(l) + 9/2 O₂(g) → 3 CO₂(g) + 4 H₂O(l)

The change in enthalpy of combustion of 1 mole of 2-propanol at 298.15 K is determined to be -2012 kJ at constant pressure.

What is ΔE for this reaction at 298K?

Enter your answer to 4 significant figures in units of kJ. Be sure to include the sign.

Answer 1

E for this reaction = -2008 kJ

Explanation

Gaseous moles of reactant = 4.5 mol (of O₂)

Gaseous moles of products = 3 mol (of CO₂)

Change in number of moles = (Gaseous moles of reactants) - (Gaseous moles of products)

Change in number of moles = (4.5 mol) - (3 mol)

Change in number of moles = 1.5 mol

Work change = (Change in number of moles) * (R) * (T)

Work change = (1.5 mol) * (8.314 J/mol-K) * (298.15 K)

Work change = 3718.2 J

Work change = 3.718 kJ

E = Q + W

E = (-2012 kJ) + (3.718 kJ)

E = -2008 kJ