Question

Manganese dioxide (MnO2(s), Delta.Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Delta.Hf = –1699.8 kJ/mol) and manganese according to the equation below.

3 upper M n upper O subscript 2 (s) plus 4 upper A l (s) right arrow 2 upper A l subscript 2 upper O subscript 3 (g) plus 3 upper M n (s).

What is the enthalpy of the reaction?
Use Delta H r x n equals the sum of delta H f of all the products minus the sum of delta H f of all the reactants..
–1,839.6 kJ
–1,179.8 kJ
1,179.8 kJ
1,839.6 kJ

Answer 1

ΔHrxn = [2(ΔHf Al2O3) + 3(ΔHf Mn)] − [3(ΔHf MnO2) + 4(ΔHf Al)]
= [2(−1699.8) + 3(0)] − [3(−520.0) + 4(0)]
= −3399.6 − (−1560.0) = −1839.6 kJ

Answer: −1,839.6 kJ