Question

Use the data in Appendix B in the textbook to calculate the following quantities. (College Chemistry II)

Part A-

N2(g)+2H2(g)?N2H4(l) (Express your answer using one decimal place.)

1. ?H? for the following reaction?

2. ?S? for this reaction?

3. From the values of ?H? and ?S?, calculate ?G? at 25 ?C?

Part B-

CH3OH(l)+O2(g)?HCO2H(l)+H2O(l) (Express your answer using one decimal place.)

1. ?H? for the following reaction?

2. ?S? for this reaction?

3. From the values of ?H? and ?S?, calculate ?G? at 25 ?C?

Please show the works, and thank you so much!

Answer 1

Solution

Part A-

1)

N₂(g) + 2H₂(g)?N₂H₄(l)

?H? for the reaction = ?(?H?)product - ?(?H?)reactants

= (+50.6 kJ/mol) – (0.0 kJ/mol + 2 x 0.0 kJ/mol)

?H? for the reaction = +50.6 kJ/mol (ANS)

2)

?S? for the reaction = ?(?S?)product - ?(?S?)reactants

= (+121.5 J/K mol) – (+191.6 J/K mol + 2 x 130.7 J/K mol)

= +121.5 J/K mol – 453 J/K mol

?S? for the reaction = -331.5 J/K mol (ANS)

3)

?G° = ?H° - T?S°

= +50600 J /mol – [(273.15 + 25)K * (-331.5 J/K mol)]

= +50600 J /mol – 98836.7 J /mol

= -48236.7 J /mol

= -48.2 J /mol (ANS)

Part B-

CH₃OH (l) + O₂ (g) ? HCO₂H (l) + H₂O (l)

1)

?H? for the reaction = ?(?H?)product - ?(?H?)reactants

= [(-425.5 kJ/mol) + (-285.83 kJ/mol)] – [(-239.03 kJ/mol) + (- 0.0 kJ/mol)]

?H° for the reaction = - 472.3 kJ/mol (ANS)

2)

?S? for the reaction = ?(?S?)product - ?(?S?)reactants

= (129.0 J/K mol + 69.91 J/K mol) – (+127.24 J/K mol + 205.3 J/K mol)

= -133.6 J/K mol

?S? for the reaction = -133.6 J/K mol (ANS)\

3)

?G° = ?H° - T?S°

= - 472300 J/mol – [(273.15 + 25)K * (-133.6 J/K mol)]

= - 472300 J/mol + 39832.84 J /mol

= -432467.2 J /mol

= -432.5 J /mol (ANS)