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Δg of carbon monoxide?
Given that:
C(s)+)2(g) --> CO2(g) ΔG=-394.4 kJ
2CO(g)+O2(g) --> 2 CO2(g) ΔG=-514.4 kJ
calculate ΔG of carbon monoxide.
C(s) + 1/2 O2(g) --> CO(g)
1 Answer
- GeorgeSiO2Lv 71 decade agoFavorite Answer
The first law of thermodynamics allows you to manipulate ΔG equations just like regular algebraic eqns.
C(s)+O2(g) → CO2(g) (A) ΔG = -394.4 kJ
2CO(g)+O2(g) → 2 CO2(g) (B) ΔG = -514.4 kJ
C(s) + 1/2 O2(g) → CO(g) ΔG?
We see that we have to have CO on the RHS so let's reverse equation B which reverses the sign of ΔG and take half of the value.
CO2(g) → CO(g) + 0.5O2(g) ΔG = +(514.4)/2 = +257.2 kJ
Add this to Eqn A
C(s)+O2(g) + CO2(g) → CO2(g) + CO(g) + 0.5O2(g) ΔG = 257.2 - 394.4
Cancel CO2 on both sides of eqn and subtract the 0.5O2 (LHS) from 1.0 O2(RHS):
C(s) + 0.5O2(g) → CO(g) ΔG = -137.2 kJ mol^-1
Lit Value (2011 CRC Handbook) Sect. 5.20 = -137.2 kJ mol^-1 (Wow, that was lucky!)