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QUESTION 4

The overall enthalpy of a reaction can be determined using Hess’s law to add up what quantities?

The enthalpy of formation reactions.

The specific heat of each reactant.

The specific heat of each product.

The heat of vaporization of each reactant.
3 points
QUESTION 5

Which of these reactions release heat?

SOCl2(l) + H2O(l) → SO2(g) + 2 HCl(g) ΔH = +10.3 kJ

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

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

Cl3PO(g) → PCl3(g) + 1/2 O2(g) ΔH = +285.7 kJ
4 points
QUESTION 6

To calculate the enthalpy of the reaction, N2 (g) + 3H2 (g) → 2NH3 (g), you need to subtract which heats of formation?

N2 (g) and H2 (g)

NH3 (g)

N2 (g), H2 (g), and NH3(g)

none of them
3 points
QUESTION 7

Using the table below, what is the change in enthalpy for the following reaction when all reactants and products are in the gaseous state?

Nitrogen Dioxide → Nitrogen Monoxide + Oxygen

Substance ΔHf (kJ/mol)
NO2 (g) 33.85
NO (g) 90.37
O2 (g) 0.0
O3 (g) 142.0


113.0 kJ

255.0 kJ

56.5 kJ

-113.0 kJ
3 points
QUESTION 8

If all the coefficients of a reaction are multiplied by 1 during a Hess’s law calculation, what must be done to the ΔH for that reaction?

It must be divided by 2.

It must be multiplied by 3.

It must have 1 subtracted from it.

It must be left alone.

Respuesta :

Nobelium
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."
jufsanabriasa

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!