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Charra [1.4K]
3 years ago
11

A reaction that occurs in the internal combustion engine is n2(g) + o2(g) ⇌ 2 no(g) (a) calculate δh o and δs o for the reaction

at 298 k. δh o = kj δs o = j/k (b) assuming that these values are relatively independent of temperature, calculate δg o at 55°c, 2570°c, and 3610°c. 55°c: kj 2570°c: kj 3610°c: kj
Chemistry
1 answer:
jekas [21]3 years ago
7 0
1) ΔrH = 2mol·ΔfH(NO) - (ΔfH(O₂) + ΔfH(N₂)).
ΔrH = 2 mol · 90.3 kJ/mol - (0 kJ/mol + 0 kJ/mol).
ΔrH = 180.6 kJ.
2) ΔS = 2mol·ΔS(NO) - (ΔS(O₂) + ΔS(N₂)).
ΔS = 2mol · 210.65 J/mol·K - (1mol · 205 J/mol·K + 1 mol · 191.5 J/K·mol).
ΔS = 24.8 J/K.
3) ΔG = ΔH - TΔS.
55°C: ΔG = 180.6 kJ - 328.15 K · 24.8 J/K = 172.46 kJ.
2570°C: ΔG = 180.6 kJ - 2843.15 K · 24.8 J/K = 110.09 kJ.
3610°C: ΔG = 180.6 kJ - 3883.15 K · 24.8 J/K = 84.29 kJ.
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Answer:

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Explanation:

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3 years ago
Consider the reaction of gaseous hydrogen with gaseous oxygen to produce gaseous water. Given that the first picture represents
Bogdan [553]

The question is incomplete. There's missing the image, which is shown below.

Answer:

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Explanation:

The reaction between hydrogen gas and oxygen gas to form water is:

2H₂(g) + O₂(g) → 2H₂O(g)

So, for 1 mol of O₂ is necessary 2 moles of H₂ form 2 moles of H₂O. As the images below there's 8 molecules of H₂, 4 molecules of O₂, 12 molecules in the mixture, and 8 molecules of H₂O. Thus, there are stoichiometric values.

All the images are at the same temperature and pressure, so, by the ideal gas law:

PV= nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

The number of moles and molecules are related, so let's substitute it in the equation. For the H₂:

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Answer:

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Explanation:

Given parameters:

Formula of the compound:

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Unknown:

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Solution:

The number of moles of any substance is given as:

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