[standard enthalpy of formation]

remember to show work and explain.

3. from this data, calculate the change in enthalpy for:

4. use the information below to solve for the ∆h for the reaction:

3. ΔH = 0.30 kJ; 4. ΔH = -84.6 kJ

Step-by-step explanation:

Question 3:

We have three equations:  

(I)  S(r) + O₂ → SO₂;         ΔH = -296.06 kJ

(II) S(m) + O₂ ⟶ SO₂;     ΔH = -296.36 kJ

From these, we must devise the target equation:  

(III) S(r) ⟶ S(m);              ΔH = ?  

The target equation has 1S(r) on the left, so you rewrite Equation(I).

(IV) S(r) + O₂ ⟶ SO₂;      ΔH = -296.06 kJ  

Equation (IV) has 1SO₂ on the right, and that is not in the target equation.  

You need an equation with 1SO₂ on the left, so you reverse Equation (II).  

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

(V)  SO₂ ⟶ S(m) + O₂ ;     ΔH = 296.36 kJ

Now, you add equations (IV) and (V), 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 (III)

(IV) S(r) + O₂ ⟶ SO₂;       ΔH = -296.06 kJ  

(V)  SO₂ ⟶ S(m) + O₂;      ΔH =  296.36 kJ

(III) S(r) ⟶ S(m);                ΔH =       0.30 kJ

Question 4  

We have three equations:  

(I)  C + O₂ ⟶ CO₂;                                ΔH = -393.5 kJ

(II) H₂ + ½O₂ ⟶ H₂O;                           ΔH = -285.8 kJ

(III) 2C₂H₆ + 7O₂ ⟶ 4CO₂ + 6H₂O;     ΔH = -296.8 kJ

From these, we must devise the target equation:  

(IV) 2C + 3H₂ → C₂H₆; ΔH = ?  

The target equation has 2C on the left, so you double Equation(I).

When you double an equation, you double its ΔH.

(V) 2C + 2O₂ ⟶ 2CO₂;                          ΔH = -787.0 kJ

Equation (V) has 2CO₂ on the right, and that is not in the target equation.  

You need an equation with 2CO₂ on the left, so you reverse Equation (III) and divide it by 2.

(VI) 2CO₂ + 3H₂O ⟶ C₂H₆ + ⁷/₂O₂;     ΔH = 148.4 kJ

Equation (V) has 3H₂O on the left, and that is not in the target equation.

You need an equation with 3H₂O on the right. Triple Equation (III).

(VII) 3H₂ + ³/₂O₂ ⟶ 3H₂O;                   ΔH = -857.4 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)     2C + 2O₂ ⟶ 2CO₂;                      ΔH =  -787.0 kJ

(VI)   2CO₂ + 3H₂O ⟶ C₂H₆ + ⁷/₂O₂;     ΔH = 1559.8 kJ

(VII) 3H₂ + ³/₂O₂ ⟶ 3H₂O;                     ΔH = - 857.4 kJ

(IV)  2C + 3H₂ → C₂H₆;                            ΔH =    -84.6 kJ

answer: i would say b.

explanation: