Answer to Question #119310 in Organic Chemistry for emily

Question

Calculate the enthalpy of the following reaction: (3 marks) 2Mg + O2 → 2MgO Given: Mg + 2HCl → MgCl2 + H2 ΔH = -21.0 kJ MgO + 2HCl → MgCl2 + H2 O ΔH = -33.5 kJ H2 + ½ O2 → H2 O ΔH = -76.0 kJ

Answer 1

Solution:

Given:

(1) Mg + 2HCl → MgCl₂ + H₂, ΔH₁ = -21.0 kJ

(2) MgO + 2HCl → MgCl₂ + H₂O, ΔH₂ = -33.5 kJ

(3) H₂ + ½O₂ → H₂O, ΔH₃ = -76.0 kJ

2Mg + O₂ → 2MgO ΔH_r = ???

Using Hess's law to calculate the reaction enthalpy:

ΔH_r = 2×ΔH₁ - 2×ΔH₂ + 2×ΔH₃

ΔH_r = 2×(-21.0 kJ) - 2×(-33.5 kJ) + 2×(-76.0 kL) = -127 kJ

ΔH_r = -127 kJ

Answer: The enthalpy (ΔH_r) of this reaction is -127 kJ.

NB! The values of reaction enthalpies (1), (2) and (3) are not correct!

(1) Mg + 2HCl → MgCl₂ + H₂, ΔH₁ = -427.99 kJ/mol

(2) MgO + 2HCl → MgCl₂ + H₂O, ΔH₂ = -74.66 kJ/mol

(3) H₂ + ½O₂ → H₂O, ΔH₃ = -285.5 kJ/mol

Then, ΔH_r = -1277.66 kJ/mol