# Repeat above calculation but with p co 2 5 bar δ r g

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Repeat above calculation but with p CO 2 = 5 bar. Δ r G m = Δ r G m + RT ln Q = 8 . 4 kJ / mol + (8 . 314 472 × 10 - 3 kJ K - 1 mol - 1 )(298 . 15 K) × ln 0 . 1 5 = - 1 . 3 kJ / mol Conclusion: Δ r G m < 0 therefore carbonation is thermodynamically allowed under these conditions.

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Reaction of lead(II) nitrate with ammonium sulfate I 20 mL of a 0.040 mol/L solution of lead(II) nitrate is mixed with 15 mL of a 0.0031 mol/L solution of ammonium sulfate at 25 C. Does a precipitate form? Answer: Δ r G m = - 18 . 9 kJ / mol < 0 therefore a precipitate forms.
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