How to solve this delta G problem (gen chem 2)

Question

2. Consider the following reaction, for which data at 25oC are given:

2Na+(aq) + CO32-(aq) → Na2CO3(s)

∆Hof (kJ/mol) -240.1 -677.1 -1130.8

So (J/K mol) 59.1 -56.9 138.8

∆Gof (kJ/mol) -261.9 -527.9 -1048.0

Calculate ∆Go of this reaction in kJ at a temperature of 40 C.

*** Show your work.

A) +258kJ

B) -17.8kJ

C) -258kJ

D) -0.723kJ

E) +3.04 x 103kJ

F) +5.99 x 104kJ

G) +2.24kJ

why would the answer be G? i have the answer key just dont know how to get there

Answer 1

Use ∆Gº = ∆Hº - T∆Sº

To calculate ∆Hº and ∆Sº, we will use ∑products - ∑reactants, as follows:

∆Hº = -1130.8 - (2*-240.1 + -677.1) = -1130.8 - (-1157.3) = +26.5 kJ/mole

∆Sº = 138.8 - (2*59.1 -56.9) = 138.8 - 61.3 = 77.5 J/Kmole = 0.0775 kJ/Kmol (change units to kJ/Kmol)

Plug these values into ∆Gº = ∆Hº - T∆Sº where T = 40ºC + 273 = 313K

∆Gº = 26.5 - (313)(0.0775) = 26.5 - 24.26 = +2.24 kJ / mole (3 sig.figs.)