Calculate ΔG∘ at 298 K for the following reactions.Answers
Answer:
![AG = AG° + RT In [C][D] [A][B]](https://i.imgur.com/JOTHOwc.png)
At equilibrium = Det G= 0
![At equilibrium condition, ago 49°= -Rt lui [C] ?] TAB TEJ = -KT In .. (als) + (02(g) + + O2(9) CO₂ (s) : Q = [ Ca.(03) - [la](https://i.imgur.com/Z074UHA.jpeg)
Let's consider the following reversible reaction: A+B= C+D The following equation relates the standard-state free energy of reaction with the free energy at any point in a given reaction (not necessarily at standard-state conditions): AG = AGⓇ + RT In Q • AG = free energy at any moment • AG° = standard-state free energy • Ris the ideal gas constant = 8.314 J/mol-K • Tis the absolute temperature (Kelvin) • In Q is natural logarithm of the reaction quotient
AG = AG° + RT In [C][D] [A][B]
At equilibrium condition, ago 49°= -Rt lui [C] ?] "TAB TEJ = -KT In .. (als) + (02(g) + + O2(9) CO₂ (s) : Q = [ Ca.(03) - [la] [uaj [02] as [ solid] 21: 8. I 1772 .= -8.314 3/6.1.K • 298K .An Trogloszyn AGO = 2 477 574 h [6] to Paste8: Caloz is) →6019 +102 (9) Q = [60] [(00)Los ] [16 (03] 46° = - RT ln [co2] Rhod flaco = 46° = -8:319X2984. In P602) amo) =-2477.57 In [002] Olmo).
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