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Wittaler [7]
3 years ago
6

Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this li

st of thermodynamic properties.HCl(g)+NaOH(s)⟶NaCl(s)+H2O(l)
Chemistry
1 answer:
Anastasy [175]3 years ago
7 0

Answer:

-179.06 kJ

Explanation:

Let's consider the following balanced reaction.

HCl(g) + NaOH(s) ⟶ NaCl(s) + H₂O(l)

We can calculate the standard enthalpy change for the reaction (ΔH°r) using the following expression.

ΔH°r = 1 mol × ΔH°f(NaCl(s)) + 1 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(HCl(g)) - 1 mol × ΔH°f(NaOH(s))

ΔH°r = 1 mol × (-411.15 kJ/mol) + 1 mol × (-285.83 kJ/mol) - 1 mol × (-92.31 kJ/mol) - 1 mol × (-425.61 kJ/mol)

ΔH°r = -179.06 kJ

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Angelina_Jolie [31]

Answer:

The structure with the ring flipped is the most stable

Explanation:

We have the  trans 1,2 - dimethylcyclohexane. With the wedge/dash structure we could not figure is this form is stable (If we do a comparison with the cis structure). But when we do a chair structure and ring flipped structure, this is easier to look.

The picture attached shows the structures, they are labeled as 1, 2 and 3, according to this problem.

In the chair structure, according to the picture below, you can see that both methyls are heading in the axial positions of the ring (One facing upward and the other downward). This is pretty stable, however, when the methyls are in those positions, the methyl position 1, can undergoes an 1,3 diaxial interactions with the hydrogens atoms (They are not drawn, but still are there), so this interaction makes this structure a little less stable that it can be.

On the other side, the ring flipped structure, we can see that both methyls are in the equatorials positions of the ring, and in these positions, it can avoid the 1,4 diaxial interactions with the hydrogens atoms, making this structure the most stable structure.

Hope this helps

6 0
3 years ago
A certain reaction is thermodynamically favored at temperatures below 400. K, but it is not favored at temperatures above 400. K
Ilya [14]

Answer:

-0.050 kJ/mol.K

Explanation:

  • A certain reaction is thermodynamically favored at temperatures below 400. K, that is, ΔG° < 0 below 400. K
  • The reaction is not favored at temperatures above 400. K, that is. ΔG° > 0 above 400. K

All in all, ΔG° = 0 at 400. K.

We can find ΔS° using the following expression.

ΔG° = ΔH° - T.ΔS°

0 = -20 kJ/mol - 400. K .ΔS°

ΔS° = -0.050 kJ/mol.K

8 0
3 years ago
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Alik [6]

Answer:

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Hope this helps :-)

Have a great rest of your day or night!

Enjoy your studies and assignments

<3 simplysun

ps. I do not own any of these answers so please don't give full credit to me

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