contestada

Given the following data:
N2(g)2NO(g)++O2(g)O2(g)2N2O(g)→→→2NO(g),2NO2(g),2N2(g)+O2(g),ΔH=+180.7kJΔH=−113.1kJΔH=−163.2kJ
use Hess's law to calculate ΔH for the following reaction:
N2O(g)+NO2(g)→3NO(g)

Respuesta :

Dejanras

Answer is: enthalpy is 155.65 kJ.

Reaction 1: N₂(g) + O₂(g) → 2NO(g); ΔrH₁ = 180.7 kJ.

Reaction 2: 2NO(g) + O₂(g) → 2NO₂(g); ΔrH₂ = -113.1 kJ.

Reaction 3: 2N₂O → 2N₂(g) + O₂(g); ∆rH₃ = -163.2 kJ.

Reaction 4: N₂O(g) + NO₂(g) → 3NO(g); ΔrH₄ = ?.

Using Hess's law reaction number 4 is sum of reaction number 1 and half of reaction number 3 minus half of the reaction number 2.

ΔrH₄ = ΔrH₁ + 1/2ΔrH₃ - 1/2ΔrH₂.

ΔrH₄ = 180.7 kJ + 1/2 · (-163.2 kJ) - 1/2 · (-113.1 kJ).

ΔrH₄ = 155.65 kJ.

Hess's law states that the enthalpy change in the chemical reaction is same, whether the reaction takes place in one or several steps. For the given equation, the enthalpy change of overall reaction is 155. 65 kJ.

[tex]\begin{aligned}\text{N}_2(\text{g})2\text{NO}(\text{g})+\text{O}_2(\text{g})\text{O}_2(\text{g})2\text{N}_2\text{O}(\text{g})\rightarrow2\text{NO}(\text{g}),2\text{NO}_2(\text{g}),2\text{N}_2(\text{g})+\text{O}_2(\text{g})\end{aligned}[/tex]

Given that,

1. The enthalpy change for the reaction between nitrogen and oxygen to form nitrogen oxide is:

  • [tex]\begin{aligned}\text{N}_2(\text{g})+\text{O}_2(\text{g})\rightarrow2\text{NO}(\text{g});\Delta\text{rH}_1&=180.7\text{kJ}\end{aligned}[/tex]

2. The enthalpy change in the reaction between nitrogen oxide and oxygen to form nitrogen dioxide will be:

  • [tex]\begin{aligned}2\text{NO}(\text{g})+\text{O}_2(\text{g})\rightarrow2\text{NO}_2(\text{g});\Delta\text{rH}_2&=-113.1\text{kJ}\end{aligned}[/tex]

3.  The enthalpy change in breakdown of nitrous oxide to form nitrogen and oxygen:

  • [tex]\begin{aligned}2\text{N}_2\text{O}\rightarrow2\text{N}_2(\text{g})+\text{O}_2(\text{g});\Delta\text{rH}_3&=-163.2\text{kJ}\end{aligned}[/tex]

4. The reaction between nitrous oxide and nitrogen dioxide will from nitrogen oxide as:

  • [tex]\begin{aligned}\text{N}_2\text{O}(\text{g})+\text{NO}_2(\text{g})\rightarrow3\text{NO}(\text{g});\Delta\text{rH}_4&=?\end{aligned}[/tex]

Using Hess's law on the fourth reaction, we get sum of first reaction and halved the enthalpy change of second and third reaction.

[tex]\begin{aligned}\Delta\text{rH}_4&=\Delta\text{rH}_1+\frac{1}{2}\Delta\text{rH}_3-\frac{1}{2}\Delta\text{rH}_2\\&=180.7\text{kJ}+\frac{1}{2}\times(-163.2\text{kJ})-\frac{1}{2}\times(-113.1\text{kJ})\\&=155.65\text{kJ}\end{aligned}[/tex]

Therefore, the enthalpy change for the equation will be 155.65 kJ.

To know more about Hess's law, refer to the following link:

https://brainly.com/question/14469426?referrer=searchResults