The rate constant for the decomposition of the AB molecule at 27.0 degree C is 0
ID: 539549 • Letter: T
Question
The rate constant for the decomposition of the AB molecule at 27.0 degree C is 0.0641 s^-1. Calculate the rate constant at 89.0 degree C if the activation energy is 17.2 kJ/mol. For the following reaction at 25 degree C, Delta H = -39.61 kJ/mole. N_2O_5(g) + H_2O_(g) rightarrow 2 HNO_3(g) If the absolute entropies of N_2O_5 (g). H_2O (g) and HNO_3 (g) are 356 J/mole.K, 188.7 J/mole.K and 266.2 J/mole.K, respectively, for what temperature range is the reaction spontaneous? [Assume Delta H and Delta S are constant with changes in temperature.)Explanation / Answer
6)
T1 = 27.0 oC
=(27.0+273)K
= 300.0 K
T2 = 89.0 oC
=(89.0+273)K
= 362.0 K
K1 = 6.41*10^-2 s-1
Ea = 17.2 KJ/mol
= 17200 J/mol
use:
ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)
ln(K2/6.41*10^-2) = (17200/8.314)*(1/300.0 - 1/362.0)
ln(K2/6.41*10^-2) = 2069*(5.709*10^-4)
K2 = 0.209 s-1
Answer: 0.209 s-1
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