The standard free energy of activation of a reaction A is 73.4 kJ mol–1 (17.5 kcal mol–1) at 298 K. Reaction B is one 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. (a) What is the standard free energy of activation of reaction B?
a.) What is the standard free energy of activation of reaction B?
B.) What is the standard free energy of activation of the reverse of reaction A?
C.)What is the standard free energy of activation of the reverse of reaction B?
I would do this.
ln (k2/k1) = Ea/RT
You know k2 is 1E6k1 and you know Ea for A, solve for Ea for B in which k1 is the rate constant for reaction A and k2 is the constant for reaction B.
If products are more stable than reactants then reactions A and B are exothermic and Ea reverse is Ea + 10 kJ/mol.
Check out my thinking before you do anything.
a) Well, if reaction B is one million times faster than reaction A, we can assume it's supercharged like The Flash on caffeine. So, let's calculate the standard free energy of activation for reaction B by dividing the standard free energy of activation of reaction A (73.4 kJ mol–1) by one million. That's like dividing a cheetah by a snail. The result is 0.0734 kJ mol–1, or you can think of it as Usain Bolt giving a piggyback ride to a turtle. So, the standard free energy of activation of reaction B is approximately 0.0734 kJ mol–1.
b) To find the standard free energy of activation of the reverse of reaction A, we just need to flip the sign. It's like turning an "Oops" into a "S'poop" or a "Yikes" into a "Sehkiy." So, the standard free energy of activation of the reverse of reaction A is -73.4 kJ mol–1 (or -17.5 kcal mol–1).
c) Now, for the standard free energy of activation of the reverse of reaction B, we can use the same strategy as before. Since the reverse of reaction B is just the reverse of reaction A but supercharged, we divide the standard free energy of activation of reaction B (0.0734 kJ mol–1) by one million, just like dividing a lightning bolt by a sloth on a lazy day. The result is approximately 0.0000000734 kJ mol–1, or you can think of it as a gummy bear outrunning a snail in a marathon. So, the standard free energy of activation of the reverse of reaction B is approximately 0.0000000734 kJ mol–1.
a) To determine the standard free energy of activation for reaction B, we can use the relationship between reaction rates and activation energies. Since reaction B is one million times faster than reaction A at the same temperature, we can say that the rate constant (k) for reaction B is 1 million times greater than the rate constant for reaction A.
The relationship between the rate constant (k) and the standard free energy of activation (ΔG^‡) is given by the Arrhenius equation:
k = Ae^(-ΔG^‡/RT)
Where:
k = rate constant
A = pre-exponential factor (frequency factor)
ΔG^‡ = standard free energy of activation
R = gas constant (8.314 J/(mol*K))
T = temperature in Kelvin
Since both reactions are at the same temperature (298 K) and the pre-exponential factor (A) is likely to be similar for both reactions, we can simplify the equation to:
k_B = 1 million * k_A
e^(-ΔG^‡_B/RT) = 1 million * e^(-ΔG^‡_A/RT)
Taking the natural logarithm of both sides:
-ln(1 million) = (-ΔG^‡_B/RT) - (-ΔG^‡_A/RT)
Evaluating the natural logarithm:
-ln(1 million) = (-ΔG^‡_B/RT) - (-ΔG^‡_A/RT)
13.8 = (-ΔG^‡_B/RT) - (-73.4/RT)
13.8 = (-ΔG^‡_B + 73.4)/RT
Since temperature (T) is constant:
13.8 * R = -ΔG^‡_B + 73.4
Rearranging the equation:
ΔG^‡_B = 73.4 - 13.8 * R
Using the value of R = 8.314 J/(mol*K), we can compute:
ΔG^‡_B = 73.4 - 13.8 * 8.314
ΔG^‡_B = 73.4 - 114.69132
ΔG^‡_B ≈ -41.291 J/mol
Therefore, the standard free energy of activation of reaction B is approximately -41.291 J/mol.
b) The standard free energy of activation of the reverse of reaction A can be determined by simply reversing the sign of the activation energy for reaction A.
Therefore, the standard free energy of activation for the reverse of reaction A is:
ΔG^‡_reverse_A = -ΔG^‡_A = -73.4 kJ/mol.
c) Similarly, the standard free energy of activation for the reverse of reaction B can be determined by reversing the sign of the activation energy for reaction B:
ΔG^‡_reverse_B = -ΔG^‡_B = -(-41.291) J/mol = 41.291 J/mol.
Therefore, the standard free energy of activation for the reverse of reaction B is approximately 41.291 J/mol.
To determine the standard free energy of activation of reaction B, we can use the given information. We know that reaction B is one million times faster than reaction A at the same temperature.
The rate constant of a reaction is related to the standard free energy of activation (ΔG‡) through the Arrhenius equation:
k = Ae^(-ΔG‡/RT)
Where k is the rate constant, A is the pre-exponential factor, R is the gas constant, and T is the temperature in Kelvin.
Since reaction B is one million times faster than reaction A, we can write:
k_B = 1,000,000 * k_A
Now, let's compare the rate constants using the Arrhenius equation:
Ae^(-ΔG‡_B/RT) = 1,000,000 * Ae^(-ΔG‡_A/RT)
The exponential term (e^(-ΔG‡/RT)) cancels on both sides, giving us:
ΔG‡_B = ΔG‡_A + RT * ln(1,000,000)
Now, we can substitute the given values into the equation:
ΔG‡_A = 73.4 kJ mol^(-1) = 17.5 kcal mol^(-1)
R = 8.314 J mol^(-1) K^(-1)
T = 298 K
To calculate ln(1,000,000), we take the natural logarithm of 1,000,000:
ln(1,000,000) ≈ 13.82
Now, we can substitute the values into the equation:
ΔG‡_B = 17.5 kcal mol^(-1) + (8.314 J mol^(-1) K^(-1) * 298 K * ln(1,000,000))
Calculating the expression on the right-hand side:
ΔG‡_B ≈ 17.5 kcal mol^(-1) + (8.314 J mol^(-1) K^(-1) * 298 K * 13.82)
ΔG‡_B ≈ 17.5 kcal mol^(-1) + 32081.65 J mol^(-1)
Converting J to kcal:
ΔG‡_B ≈ 17.5 kcal mol^(-1) + (32081.65 J mol^(-1) / 4184 J kcal^(-1))
ΔG‡_B ≈ 17.5 kcal mol^(-1) + 7.67 kcal mol^(-1)
ΔG‡_B ≈ 25.2 kcal mol^(-1)
Therefore, the standard free energy of activation of reaction B is approximately 25.2 kcal mol^(-1).