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(10 pts) A chemist is investigating a reaction that has a ΔG。f-5.5 kJ/mol. a) At equilibrium what percentage of the reaction mixture will be products? Show your work. b) Will equilibrium be achieved quickly, slowly, or do you need more information to make that determination? Explain your reasoning
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Answer #1

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dG = -RT*ln(K)

Assume T = 298K, since none is given

K = epx(-dG/(RT))

K = exp(5500/8.314/298)

K = 9.206

assume A --> B so

K = B/A = products/Reactants

9.206 = Products/Reactants

Products = 9.206 *Reactants

assume reactants = 1, then P = 9.206

%P = 9.206 /(1+9.206 ) * 100 = 90.2% are products, then 9.8% are reactants

b)

No information given, dG data only state equilbirium, but no kinetic/rate of reaction

we will need activation energy or any other relevant data in equilbirium


answered by: ANURANJAN SARSAM
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Answer #2

Apply

dG = -RT*ln(K)

Assume T = 298K, since none is given

K = epx(-dG/(RT))

K = exp(5500/8.314/298)

K = 9.206

assume A --> B so

K = B/A = products/Reactants

9.206 = Products/Reactants

Products = 9.206 *Reactants

assume reactants = 1, then P = 9.206

%P = 9.206 /(1+9.206 ) * 100 = 90.2% are products, then 9.8% are reactants

b)

No information given, dG data only state equilbirium, but no kinetic/rate of reaction

we will need activation energy or any other relevant data in equilbirium

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