Question

Consider the following reaction at constant P. Use the information here to determine the value of ΔSsurr at 355 K. Predict whether or not this reaction will be spontaneous at this temperature. 2NO(g) + O2(g) → 2NO2(g) ΔH = -114 kJ Consider the following reaction at constant . Use the information here to determine the value of at 355 . Predict whether or not this reaction will be spontaneous at this temperature. 2(g) + (g) 2(g) = -114 ΔSsurr = +114 kJ/K, reaction is spontaneous ΔSsurr = -114 kJ/K, reaction is spontaneous ΔSsurr = -321 J/K, reaction is spontaneous ΔSsurr = +321 J/K, reaction is spontaneous ΔSsurr = +114 J/K, it is not possible to predict the spontaneity of this reaction without more information.g

Answer 1

The given reaction is as follows:

2NO (g) + O₂ (g) = 2NO₂ (g), ΔH = -114 kJ

It is known that dSsurr = -dHsys / T (Temp = 355 K)

So,  dSsurr = - (-114 × 1000) / 355

dSsurr = +321.12 J/K

Hence, the value of dSsurr is +321.12 J/K

For a reaction to be spontaneous, dG<0,

Also dStotal = dSsys + dSsurr > 0

It is known that dG = dHsys - TdSsys,

Now let us assume,

dG<0

Also, dStotal = dSsys + dSsurr > 0

(-114 × 1000) - (355 × dSsys) <0

355 × dSsys > -114 × 1000

dSsys > -321

dSsys >dSsurr

dSsys + dSsurr > 0

dStotal > 0

Thus, the assumption is correct, and the given reaction is spontaneous. Hence, the final answer is Ssurr = +321 J/K reaction is spontaneous.