Question

Must show work for questions. Please also explain, I don't understand.

11. The heat combustion of acetylene, C2H2 (g), at 25°C, is -1299 kJ/mol. At this temperature, ∆H_f° values for CO2(g) and H2O(l) are -393 and -286 kJ/mol, respectively. Calculate ∆H_f° for acetylene.

12. Consider the following data. Use Hess's law to find the change in enthalpy at 25°C for the following equation.​

Answer 1

11. 227 kJ·mol⁻¹; 12. 587 kJ

Step-by-step explanation:

11. Calculate ΔH°f

The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is

ΔH°r = ΣmΔH°f(products) - ΣnΔH°f(reactants)

2C₂H₂ + 5O₂ ⟶ 4CO₂ + 2H₂O; ΔH°c = -1299 kJ·mol⁻¹

ΔH°f/kJ·mol⁻¹: 2x 0 -393 -286

2(-1299) = 4(-393) + 2(-286) - 2x

-2598 = -1572 - 572 - 2x

-2598 = -2144 -2x

-454 = -2x

x = 227 kJ·mol⁻¹

ΔH°f for acetylene is 227 kJ·mol⁻¹

12. Hess's Law

Warning: It is almost impossible to solve a problem like this by trial and error. You must have a strategy beforehand.

We have five equations:

(I) Ca + 2C ⟶ CaC₂; ΔH = -62.8 kJ

(II) Ca + ½O₂ ⟶ CaO; ΔH = -635.5 kJ

(III) CaO + H₂O ⟶ Ca(OH)₂; ΔH = -653.1 kJ

(IV) C₂H₂ + ⁵/₂O₂ ⟶ 2CO₂ + H₂O; ΔH = -1300 kJ

(V) C + O₂ ⟶ CO₂; ΔH = -393.51 kJ

From these, we must devise the target equation:

(VI) CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂; ΔH = ?

Here comes the strategy. Remember, you can use each of the given equations only once.

Start with the target equation. It has CaC₂ on the left, so you need an equation containing CaC₂ on the left.

Reverse Equation (I). When you reverse an equation, you change the sign of its ΔH.

(VII) CaC₂ ⟶ Ca + 2C; ΔH = 62.8 kJ

Equation (VII) has Ca on the right, and that is not in the target. You need an equation with Ca on the left to cancel it. Add Equation (II)

(VIII) Ca + O₂ ⟶ CaO; ΔH = -1271.0 kJ

Equation (VII) also has 2C on the right, and there is no C in the target. We need an equation with 2C on the left to cancel it.

Double Equation (V). When you double an equation, you double its ΔH.

(IX) 2C + 2O₂ ⟶ 2CO₂; ΔH = -587.02 kJ

Now, we must eliminate the CaO from Equation (VII). Add Equation (III).

(X) CaO + H₂O ⟶ Ca(OH)₂; ΔH = -653.1 kJ

Equation (VIII) has 2CO₂ on the right. You need an equation with 2CO₂ on the left.

Reverse Equation (IV).

(XI) 2CO₂ +H₂O ⟶ C₂H₂ + ⁵/₂O₂ ; ΔH = 2600 kJ

Now, we add all five equations, cancelling substances that appear on both sides of the reaction arrow. All the formulas in italics have been cancelled

(VII) CaC₂ ⟶ Ca + 2C; ΔH = 62.8 kJ

(VIII) Ca + ½O₂ ⟶ CaO; ΔH = -635.5 kJ

(IX) 2C + 2O₂ ⟶ 2CO₂; ΔH = -787.02 kJ

(X) CaO + H₂O ⟶ Ca(OH)₂; ΔH = -653.1 kJ

(XI) 2CO₂ + H₂O ⟶ C₂H₂ + ⁵/₂O₂ ; ΔH = 2600 kJ

(VI) CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂; ΔH = 587 kJ

ΔH = 587 kJ