Question 5 Not yet answered Marked out of 1.00 Calculate the change in entropy (

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Question 5 Not yet answered Marked out of 1.00 Calculate the change in entropy (in J/K) for the following reaction at 298 K: 2N0(g) +02(g) 2NO2(g) Use the data in Appendix Il (Tro); Appendix 4 (Zumdahl) When no amounts are given, it is for the reaction as written in moles. In this case 2 moles NO and 1 Flag question mole oxygen produce 2 moles NO2.) Answer: Question 6 Not yet answered Marked out of 1.00 Calculate the change in entropy (in J/K) for the following reaction at 298 K: C(s) + H2O(g) CO(g) + H2(g) Use entropy for graphite (diamond is too expensive). Notice that the entropy of the solid is much smaller than the entropy of the gases! Answer: Flag question

Explanation / Answer

5)

we have:

Sof(NO(g)) = 210.76 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:

2 NO(g) + O2(g) ---> 2 NO2(g)

deltaSo rxn = 2*Sof(NO2(g)) - 2*Sof( NO(g)) - 1*Sof(O2(g))

deltaSo rxn = 2*(240.06) - 2*(210.76) - 1*(205.138)

deltaSo rxn = -146.538 J/K

Answer: -146.54 J/K

6)

we have:

Sof(C(s,graphite)) = 5.74 J/mol.K

Sof(H2O(g)) = 188.825 J/mol.K

Sof(CO(g)) = 197.674 J/mol.K

Sof(H2(g)) = 130.684 J/mol.K

we have the Balanced chemical equation as:

C(s,graphite) + H2O(g) ---> CO(g) + H2(g)

deltaSo rxn = 1*Sof(CO(g)) + 1*Sof(H2(g)) - 1*Sof( C(s,graphite)) - 1*Sof(H2O(g))

deltaSo rxn = 1*(197.674) + 1*(130.684) - 1*(5.74) - 1*(188.825)

deltaSo rxn = 133.8 J/K

Answer: 133.8 J/K

Feel free to comment below if you have any doubts or if this answer do not work

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