Ozone (O3) in the atmosphere can react with nitric oxide (NO): O3(g) + NO(g) --> NO2(g) +...

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Ozone (O3) in the atmosphere can react with nitric oxide (NO): O3(g) + NO(g) --> NO2(g) +...

Ozone (O₃) in the atmosphere can react with nitric oxide (NO):
O₃(g) + NO(g) --> NO₂(g) + O₂(g). ( ΔH° = –199 kJ/mol, ΔS° = –4.1 J/K·mol)

Write answers to three significant figures.

a. Calculate the ΔG°( kJ/mol) for this reaction at 25°C.

b. Determine the temperature(^oC) at which the reaction is at equilibrium.

science chemistry

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Answer 1

Answer #1

a. We know that free energy change

AGⓇ = AH-TAS

Where $\Delta H^o$ =Standard enthalpy change for reaction=-199 kJ/mol

$\Delta S^o$ =Standard entropy change for reaction=-4.1 J/K.mol

T=Temperature (K)=25°C=25+273 K=298 K

So $\Delta G^o$ =-199 kJ/mol-298 K(-4.1 J/K.mol)/1000 J/kJ (1 kJ=1000 J)

=-199 kJ/mol+1.2218 kJ/mol

=-197.7782 kJ/mol

≈-198 kJ/mol

b. At equilibrium free energy change=0

AGⓇ = 0

i.e. 0 = SVI – HV

OSVL = HV

-199 kJ/mol=Tx (-4.1 J/K.mol)

T=-199 kJ/mol/-4.1 J/K.mol

=(-199 kJ/molx1000 J/kJ)/-4.1 J/K.mol

=48536.6 K

=48536.6 K-273 K=48236.6 °C

≈4.82 x 10⁴ °C

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Answer 2

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