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Acetic anhydride + H 2 O 2 acetate G = 91.1 kJ/mol If exactly 0.5 mM acetic anhydride were added to the reaction and allowed
Acetic anhydride + H2O 2 acetate G = 91.1 kJ/mol If exactly 0.5 mM acetic anhydride were added to the reaction and allowed to form product at 37 C, how much acetate would be present at equilibrium?
Standard transformed constants are used
[H20 ] = 55.5 M
R = 8.315
The answer is 6.7*10^7 mM but can you provide an explanation please?
I believe the equation is Delta G (standard) = -R*T * ln (Product over reactants)
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