The standard free energy of activation of a reaction A is 88.6 kJ mol–1 (21.2 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.
(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?
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To answer these questions, we need to understand the concepts of standard free energy of activation and reverse reactions. Let's break down each question and explain how to get the answers.
(a) The standard free energy of activation (ΔG‡) is a measure of the energy required to initiate a reaction. In this case, we are given the ΔG‡ of reaction A as 88.6 kJ mol–1.
The question states that reaction B is one hundred million times faster than reaction A at the same temperature. The rate of a reaction is related to its activation energy, and the activation energy depends on ΔG‡. If a reaction is a hundred million times faster, it means that the activation energy of reaction B is much lower than that of reaction A.
To find the standard free energy of activation of reaction B, we need to determine the ratio between the rate constants of the two reactions. The rate constant ratio can be calculated using the Arrhenius equation:
k2 / k1 = exp((ΔG‡1 - ΔG‡2) / (R * T))
Where k1 and k2 are the rate constants for reactions A and B, ΔG‡1 and ΔG‡2 are their respective standard free energies of activation, R is the ideal gas constant, and T is the temperature in Kelvin.
Since we know that reaction B is one hundred million times faster than reaction A, the rate constant ratio can be expressed as:
k2 / k1 = 100,000,000
By plugging in the values and solving for ΔG‡2:
100,000,000 = exp((88.6 - ΔG‡2) / (R * T))
Solving this equation will give you the standard free energy of activation of reaction B.
(b) The reverse of reaction A is called the reverse reaction. The standard free energy of activation of the reverse reaction (ΔG‡R) is equal in magnitude but opposite in sign to ΔG‡ of the forward reaction.
Therefore, the standard free energy of activation of the reverse of reaction A is also 88.6 kJ mol–1.
(c) To find the standard free energy of activation of the reverse of reaction B (ΔG‡R2), we use the same concept as in (b). The standard free energy of activation of the reverse reaction is equal in magnitude but opposite in sign to ΔG‡ of the forward reaction. So, ΔG‡R2 would also be equal to the ΔG‡2 value calculated in (a) but with the opposite sign.
Note: The values provided in the question, such as the stability difference between reactants and products, are not needed to answer these specific questions but may be useful for other calculations related to the reactions.
Keep in mind that the calculations provided above require the values of ΔG‡1 and ΔG‡2, which are not given in the question. So, more information or data is needed to obtain the exact numerical values for these calculations.