mempts Score 12/11/2017 11 55 PM 412/10/2017 03-50 PM Print CalculatorPeriodic T
ID: 104644 • Letter: M
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mempts Score 12/11/2017 11 55 PM 412/10/2017 03-50 PM Print CalculatorPeriodic Table Question 9 of 15 IncorrectIncorrectIncorrect MapA The standard free energy of activation of one reaction A is 90.70 kJ mor (21.68 kcal mor). The standard free energy of activation of another reaction B is 78.70 kJ mor (18.81 kcal mol). Assume a temperature of 298 K and 1 M concentration. 90 By what factor is one reaction faster than the other? Number 1269 Which reaction is faster? O Reaction A is faster. Previous Give Up & View Solution # Try Again e Next Explanation The relationship between rate and standard free energy of activation is an exponential one. The lower the energy barrier, the faster the reaction. where R = 8.314 J/(mol-K Tis the tenuperalure in kelvins. G" is in joules, a rate, 2.3RTExplanation / Answer
(delta G0)A = 90.70 KJ/mol
= 90700 J/mol
(delta G0)B = 78.70 KJ/mol
= 78700 J/mol
use,
log(rateA/rateB) = {(delta G)B - (delta G)A}/(2.3*R*T)
log(rateA/rateB) = (90700 - 78700)/(2.3*8.314*298)
log(rateA/rateB) = 12000/5698.4
log(rateA/rateB) = 2.1058
(rateA/rateB) = 127.6
Answer : 127.6
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