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owLV2 | Online teachin' > |sjc.cengagenow.com/ilrntakeAssignm lentActivity.dolocator= assignment-take8takeAssignmentsessionLocator= assignment-take Chapter 10: Mastery 1. Compare Absolute Entropies Use the Refcrences to access important values if needed for this question. 2 Predict the Sign of Consider the reaction: N2(g) + 202(NO2(g) Using standard absolute entropies at 298K, calculate the entropy change for the system when 2.38 moles of N2(g) react at standard 1 pts 2req 4. Standard Entropy of Reaction 1 pts 2redq Questo Question Question conditions. JK 1 pts 2req 1 pts 2req 7. AG: Enthalpy, Entropy and Temper...1 pts M 80G: Calculate Temperature Limit pts 2req 9. Thermodynamics of Refining Met- pts 2req 10 AG from Free Energies of Form pts 2req Submit Answer Retry Entire Group 9 more group attempts remaining 6. AG: Predict Signs

Explanation / Answer

we have:

Sof(N2(g)) = 191.61 J/mol.K

Sof(O2(g)) = 205.138 J/mol.K

Sof(NO2(g)) = 240.06 J/mol.K

we have the Balanced chemical equation as:

N2(g) + 2 O2(g) ---> 2 NO2(g)

deltaSo rxn = 2*Sof(NO2(g)) - 1*Sof( N2(g)) - 2*Sof(O2(g))

deltaSo rxn = 2*(240.06) - 1*(191.61) - 2*(205.138)

deltaSo rxn = -121.766 J/mol.K

As per rxn, this is when 1 mol of N2 reacts

So,

for 2.38 moles of N2,

deltaSo rxn = 2.38moles * -121.766 J/mol.K

= - 290 J/K

Answer: - 290 J/K

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