The standard free energy of activation of a reaction A is 85 2 kJ mol^-1 (20 4 k
ID: 1049833 • Letter: T
Question
The standard free energy of activation of a reaction A is 85 2 kJ mol^-1 (20 4 kcal mol^-1) at 298 K Reaction B is one hundred million times faster than reaction A at the same temperature The products of each reaction are 10 0 kJ mol^-1 (2 39 kcal mol^-1) more stable than the reactants What is the standard free energy of activation of reaction B? What is the standard free of activation of the reverse of reaction A? What is the standard free energy of activation of the reverse of reaction B? The following reaction is a single-step. bimolecular reaction CH_3Br + NaOH rightarrow CH_3OH + NaBr When the concentrations of CH_3Br and NaOH are both 0 165 M, the rate of the reaction is 0 0020 M/s What is the rate of the reaction 4 the concentration of CH_3Br is doubted? What is the rate of the reaction if the concentration of NaOH is halved? What is the rate of the reaction if the concentrations of CH_3Br and NaOH are both increased by a factor of four?Explanation / Answer
Take the log of the rate ratio and then solve for G of each activation energy you need.
Therefore:
log(rate A / rate B) = GB - GA / 2.3 RT
log (1 /1x108) = GB - 85.2 kJ/mol / 2.3 (0.008314 kJ/K mol) (298K)
-8 = (GB - 85.2 kJ/mol ) / 5.69 kJ/mol
Now just solve for GB
-45.58 = GB - 85.2
GB = 39.61 KJ/mol
b.) To get the reverse just add the energy of the products to each reaction energy
GA = 85.2 KJ/mol
for reverse of reaction A = 85.2 KJ/mol + 10.0 KJ/mol
= 95.2 KJ/mol
c.) To get the reverse just add the energy of the products to each reaction energy
GB = 39.61 KJ/mol
for reverse of reaction B = 39.61 KJ/mol + 10.0 KJ/mol
= 49.61 KJ/mol
2. ) rate = K [A] [B]
[A] = CH3Br , [B] = NaOH
concentration of CH3Br = 0.165 M and NaOH = 0.165 M then rate = 0.0020 M/s
rate = K [CH3Br] [NaOH]
0.027225 K = 0.002
K = 0.07346 M-1s-1
a) rate = K [CH3Br] [NaOH]
rate = 3.999 x 10-3 M/s ~ 0.004
b) rate = K [CH3Br] [NaOH]
rate = 9.99 x 10-4 M/s ~ 0.001 M/s
c) rate = K [4 x CH3Br] [4 x NaOH]
rate = 0.03199 M/s ~ 0.032 M/s
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