A reaction that occurs in the internal combustion engine isN₂(g) + O₂(g) ⇄ 2NO(g) (a) Determine ΔH° and ΔS° for the reaction at

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1 year ago

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298 K.

1 answer:

alex41 [277] · Answer 1 · 2024-06-20T13:07:08+03:00

ΔH° and ΔS° for the reaction at 298 K is 24.8 J/mol.K and 180.58 kJ/mol.

A reaction that occurs in the internal combustion engine is

N₂(g) + O₂(g) ⇄ 2NO(g)

The standard molar entropy and enthalpy of formations can be calculated using the values of ΔS°and ΔH° as shown below.

NO, N2, and O2 have entropies of 210.65 J/mol. K, 191.5 J/mol. K, and 130.6 J/mol. K, respectively.

Fill in these values in the formula for standard entropy below.

ΔS° = 2S° NO(g) -(S° N2 (g) +S° O2(g))

ΔS°= 2 (210.65 J/mol K) - (191.5 J/mol K + 205.0 J/mol K)

= 24.8 J/mol K

Consequently, the typical entropy of reaction is 24.8 J/mol.K

NO, N2, and O2 have enthalpies of 90.29 kJ/mol, 0 kJ/mol, and 0 kJ/mol, respectively.

Replace these values in the common enthalpy formula below.

ΔH° = 2ΔH° f(NO(g)) –[ ΔH° f (N2) (g) + ΔH° f (O2)(g)]

ΔH° = 2 (90.29 kJ/mol) - ((0 kJ/mol) + ((0 kJ/mol))

= 180.58 kJ/mol

Consequently, the usual enthalpy of reaction is 180.58 kJ/mol

Learn more about ΔH° here:

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