What is the enthalpy change (in kJ) when 60.0 g of potassium metal reacts according to the following equation? Give the quantity and unit, and don't forget the + or - sign.

Question

What is the enthalpy change (in kJ) when 60.0 g of potassium metal reacts according to the following equation? Give the quantity and unit, and don't forget the + or - sign.

2K +2H2O → H2 + 2KOH +160 kJ

Answer 1

2K(s) + 2H2O(l) ==> H2(g) + 2KOH(aq) ... ∆H = +160 kJ

According to this equation, you must add 160 kJ of heat energy for every 2 moles of K that reacts. Given 60.0 g of K metal, we can now calculate the input of heat energy.

60.0 g K x 1 mol K/39.1 g x 160 kJ/2 mol K = +122.8 kJ = +123 kJ (to 3 sig. figs.)