Using the provided table, determine the enthalpy for the reaction 2 K(s) + 2 H₂O(l) → 2 KOH (aq) + H₂(g)

Question

Using the provided table, determine the enthalpy for the reaction 2 K(s) + 2 H₂O(l) → 2 KOH (aq) + H₂(g)

Table

substance ΔHf°(kJ/mol)

K(s) 0

H2O (I) -285.8

KOH (aq) -482.4

H2(g) 0

Answer 1

2K(s) + 2H2O(l) ==> 2KOH(aq) + H2(g)

0.........-285.8............-482.4...........0...........∆Hºf (kJ/mol)

Sum the ∆H of products = (2)(-482.4) + 0 = - 964.8 kJ

Sum the ∆H of reactants = 0 + (2)(-285.8) = -571.6 kJ

Now take the difference of products - reactants

-964.8 kJ - (-571.6 kJ) = - 393.2 kJ = ∆H_rxn