Question

QUESTION 4

Answer 1

Answer:

Answers are in the explanation

Explanation:

4. Hess's law states that the change of enthalpy in a chemical reaction is independent of the pathway between the initial and final states. That allows to determine the enthalpy of a reaction from the enthalpy of formation reactions

5. A reaction that release heat is an exothermic reaction. A reaction is exothermic when ΔH < 0. Thus, reaction that release heat are:

PCl₃(l) + ¹/₂ O₂(g) → POCl₃(l) ΔH = -325.7 kJ

PCl₃(g) + Cl₂(g) → PCl₅(g) ΔH = -92.5 kJ

6. Based on Hess's law the enthalpy of a reaction is the enthalpy of products - enthalpy of reactants. The heat of formation of N₂ and H₂ -by definition- is 0. Thus, you don't need to substract any heat of formation.

7. For the reaction:

2NO₂ → 2NO(g) + O₂(g)

The ΔHf is:

2ΔHf NO - 2ΔHf NO₂

2×90,37 kJ/mol - 2×33,85 kJ / mol = 113,0 kJ/mol

8. If you are multiplying by one, you aren't changing anything; therefore, the ΔH must be left alone.

I hope it helps!

Answer 2

Firstly, next time you post questions, I suggest you split the questions up so the format is easier to read.

Question 4:
If you use Hess's Law. you subtract the enthalpy of formation of the reactants from the enthalpy of formation of the products. Therefore, Question 4 is "The enthalpy of formation reactions."

Question 5:
Reactions that release heat have a negative enthalpy overall. Therefore, Question 5 is:

PCl3(l) + 1/2 O2(g) → POCl3(l) ΔH = -325.7 kJ

PCl3(g) + Cl2(g) → PCl5(g) ΔH = -92.5 kJ

Question 6:
As said in Question 4, you subtract the enthalpies of formation of the reactants. Therefore, Question 6 is N2 (g) and H2 (g).

Question 7:
Balancing the equation, we get:

2 NO2 --> O2 + 2NO

Now using Hess's Law to calculate enthalpy:

(0.0 + 2(90.37))-2(33.85)
=113.0 (Rounding using sig figs)

Therefore, Question 7 is 113.0 kJ.

Question 8:

If you are multiplying by one, you aren't changing anything; therefore, Question 8 is "It must be left alone."