Question

calculate ∆G at 290k for the following reactions : 2NO (g) +O2(g) =2NO2(g) Given ∆H = -120 KJ and ∆S= -150 J/K ls the reaction spontaneous?​

Answer 1

Answer:

We know that, ΔG=ΔH−TΔS

Given that, ΔH=−113kJmol

−1

=−113000Jmol

−1

ΔS=−145JK

−1

mol

−1

T=700K

Substituting these values in the above equation

G=−113000−700×(−145)=−11500Jmol

−1

=−11.5kJmol

−1

Answer 2

The reaction is spontaneous having a ∆G of 76.5 kJ.

Given,

Reaction-2NO(g)+O2(g)→2NO2(g)

∆H of the reaction=-120kJ

∆S of the reaction=-150J.

To find,

the ∆G of the reaction at 290K.

Solution:

• Gibb's free energy is an extensive thermodynamic property.
• It is also a state function.
• The ∆G of a reaction at constant temperature and pressure having an enthalpy ∆H and an entropy ∆S is given as:
• ∆G=∆H-T∆S.
• A reaction is said be spontaneous when its ∆G is negative.
• A reaction is said be non-spontaneous when its ∆G is positive.

The ∆G of the reaction at 290K will be:

∆G=∆H-T∆S

Converting 120 kJ in J,

∆G=(-120X1000)-290(-150)

∆G=-120000+43500

∆G=-76500 J

∆G=-76500/1000 kJ

∆G=-76.5 kJ.

As the ∆G is negative, so the reaction is spontaneous.

The ∆G of the reaction at 290K is -76.5 kJ.

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