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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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B. For the following questions, use the reaction NO2(g) N2(g) + O2(g), with ΔH = –33.1 kJ/mol and ΔS= 63.02 J/(mol·K).
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Answer:

I. Kindly, see the attached image.

II. The reaction is exothermic.

III. - 51.88 kJ/mol.

IV. The reaction is spontaneous.

Explanation:

I. Draw a possible potential energy diagram of the reaction. Label the enthalpy of the reaction.

  • Since the sign of ΔH is negative, the reaction is exothermic reaction.

In an exothermic reaction, the energy of the reactants is higher than that of the products.

<u><em>Kindly see the attached image to show you the potential energy diagram of the reaction.</em></u>

     

<em>II. Is the reaction endothermic or exothermic? Explain your answer.</em>

  • The reaction is exothermic reaction.
  • The sign of ΔH indicates wither the reaction is endothermic or exothermic one:

If the sign is positive, the reaction is endothermic.

If the sign is negative, the reaction is exothermic.

Herein, <em>ΔH = - 33.1 kJ/mol, </em>so the reaction is exothermic.

<em>III. What is the Gibbs free energy of the reaction at 25°C? </em>

∵ ΔG = ΔH - TΔS.

Where, ΔG is the Gibbs free energy change (J/mol).

ΔH is the enthalpy change (ΔH = - 33.1 kJ/mol).

T is the temperature (T = 25°C + 273 = 298 K).

ΔS is the entorpy change (ΔS = 63.02 J/mol.K = 0.06302 J/mol.K).

<em>∴ ΔG = ΔH - TΔS</em> = (- 33.1 kJ/mol) - (298 K)(0.06302 J/mol.K) = <em>- 51.88 kJ/mol.</em>

IV. Is the reaction spontaneous or nonspontaneous at 25°C?

The sign of ΔG indicates the spontaneity of the reaction:

If ΔG < 0, the reaction is spontaneous.

If ΔG = 0, the reaction is at equilibrium.

If ΔG > 0, the reaction is nonspontaneous.

Herein, <em>ΔG = - 51.88 kJ/mol, </em>so the reaction is spontaneous.

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