Question

[Standard Enthalpy of Formation]

Remember to show work and explain.

5. Calculate the change in enthalpy for:

6. Calculate the ∆H_rxn given the following


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Answer 1

this is the answer using hess’ law. you need to make the 3 equations equal the final one

Answer 2

5. ΔH = 238.7 kJ; 6 ΔH, = -1504.6 kJ

Step-by-step explanation:

Question 5:

We have three equations:

(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:

(IV) C + 2H₂ + ½ O₂ → CH₃OH; ΔH = ?

The target equation has 1C on the left, so you rewrite Equation (II)..

(V) C + O₂ ⟶ CO₂; ΔH = -393.5 kJ

Equation (V) has 1CO₂ on the right, and that is not in the target equation.

You need an equation with ½O₂ on the left, so you reverse Equation (I).

When you reverse an equation, you reverse the sign of its ΔH.

(VI) CO₂ + 2H₂O ⟶ CH₃OH + ³/₂O₂; ΔH = 726.4 kJ

Equation (VI) has 2H₂O on the left, and that is not in the target equation.

You need an equation with 2H₂O on the right. Double Equation (III).

When you double an equation, you double its ΔH.

(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)

(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 + 2H₂ + ½ O₂ → CH₃OH; ΔH = -238.7 kJ

Question 6

We have three equations:

(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:

(IV) 3Mg + SO₂ → MgS + 2MgO; ΔH = ?

The target equation has 1SO₂ 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.

You need an equation with 1S on the left, so you rewrite Equation (II).

(VI) Mg + S ⟶ MgS; ΔH = -598.0 kJ

Equation (V) has 1O₂ on the right, and that is not in the target equation.

You need an equation with 1O₂ on the left. Double Equation (I).

(VII) 2Mg + O₂⟶ 2MgO ; ΔH = -1203.4 kJ

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)

(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