calculate ΔH° for the equation

Question

HNO 3 (aq) ΔH°f = –207 kJ/mol

Mg(OH) 2 (s) ΔH°f = –925 kJ/mol

Mg(NO 3 ) 2 (aq) ΔH°f = –875 kJ/mol

H 2 O(l) ΔH°f = –285.8 kJ/mol

(a) Given the ΔH°f values above, calculate ΔH° for the equation below (5 marks):

2HNO 3 (aq) + Mg(OH) 2 (s) → Mg(NO 3 ) 2 (aq) + 2H 2 O(l)

(b) Is the reaction above exothermic or endothermic? How can you tell?

Answer 1

Δ H for the reaction = sum of heats of formation of products – sum of heats of formation of reactants, each multiplied by the coefficients in the balanced equation.

Δ H = (-875 – 2x -285.8) – (-2 x 207 – 925) kj = - 516 kj.

The reaction is exothermic as ΔH is negative.

check math.