5. ΔH = 238.7 kJ; 6 ΔH, = -1504.6 kJ
Step-by-step explanation:
Question 5:
We have three equations: Â
(I)  CH₃OH + ³/₂O₂ ⟶ CO₂ + 2H₂O;  ΔH = -726.4 kJ
(II)  C + O₂ ⟶ CO₂;                ΔH = -393.5 kJ
(III) H₂ + ½O₂ ⟶ H₂O;             ΔH = -285.8 kJ
From these, we must devise the target equation: Â
(IV) C + 2Hâ‚‚ + ½ Oâ‚‚ → CH₃OH; ΔH = ? Â
The target equation has 1C on the left, so you rewrite Equation (II).. Â
(V) C + O₂ ⟶ CO₂;                ΔH = -393.5 kJ
Equation (V) has 1COâ‚‚ on the right, and that is not in the target equation. Â
You need an equation with ½Oâ‚‚ on the left, so you reverse Equation (I). Â
When you reverse an equation, you reverse the sign of its ΔH. Â
(VI) CO₂ + 2H₂O ⟶ CH₃OH + ³/₂O₂;  ΔH = 726.4 kJ
Equation (VI) has 2Hâ‚‚O on the left, and that is not in the target equation.
You need an equation with 2Hâ‚‚O on the right. Double Equation (III).
When you double an equation, you double its ΔH. Â
(VII) 2H₂ + O₂ ⟶ 2H₂O;             ΔH = -571.6 kJ
Now, you add equations (V), (VI), and (VII), cancelling species that appear on opposite sides of the reaction arrows.
When you add equations, you add their ΔH values.
We get the target equation (IV) Â
(V)  C + O₂ ⟶ CO₂;                ΔH = -393.5 kJ
(VI)  CO₂ + 2H₂O ⟶ CH₃OH + ³/₂O₂ ;  ΔH =  726.4 kJ
(VII) 2H₂ + O₂ ⟶ 2H₂O;             ΔH = -571.6 kJ
(IV)  C + 2Hâ‚‚ + ½ Oâ‚‚ → CH₃OH;       ΔH = -238.7 kJ Â
Question 6
We have three equations: Â
(I) Mg + ½O₂ → MgO;  ΔH = -601.7 kJ
(II) Mg + S ⟶ MgS;    ΔH = -598.0 kJ
(III) S + O₂ ⟶ SO₂;    ΔH = -296.8 kJ
From these, we must devise the target equation: Â
(IV) 3Mg + SOâ‚‚ → MgS + 2MgO; ΔH = ? Â
The target equation has 1SOâ‚‚ on the left, so you reverse Equation(III).
(V) SO₂ ⟶ S + O₂;     ΔH = 296.8 kJ
Equation (V) has 1S on the right, and that is not in the target equation. Â
You need an equation with 1S on the left, so you rewrite Equation (II). Â
(VI) Mg + S ⟶ MgS;     ΔH = -598.0 kJ
Equation (V) has 1Oâ‚‚ on the right, and that is not in the target equation.
You need an equation with 1Oâ‚‚ on the left. Double Equation (I).
(VII) 2Mg + O₂⟶ 2MgO ;  ΔH = -1203.4 kJ Â
Now, you add equations (V), (VI), and (VII), cancelling species that appear on opposite sides of the reaction arrows.
We get the target equation (IV) Â
(V)  SO₂ ⟶ S + O₂;           ΔH =   296.8 kJ
(VI)  Mg + S ⟶ MgS;          ΔH = -  598.0 kJ
(VII) 2Mg + O₂ ⟶ 2MgO;       ΔH = -1203.4 kJ
(IV)  3Mg + SO₂ → MgS + 2MgO; ΔH = -1504.6 kJ