In an exothermic reaction, delta H was found to be -40 kJ, and the activation energy to be 30 kJ. What would be the activation energy of the reverse reaction?
Activation energy reverse=30kJ-(-40kJ)=70kJ
To find the activation energy of the reverse reaction, we need to understand the concept of activation energy and its relationship to the forward and reverse reactions in a chemical process.
The activation energy (Ea) is the minimum energy required for a chemical reaction to occur. In an exothermic reaction, the forward reaction releases energy, resulting in a negative delta H, like in this case where delta H is -40 kJ. The reverse reaction, on the other hand, absorbs energy, requiring an input of energy.
Now, the relationship between the activation energy of the forward reaction (Ea forward) and the activation energy of the reverse reaction (Ea reverse) can be explained using the concept of the Gibbs free energy of reaction (∆G). According to the Arrhenius equation:
∆G = ∆H - T∆S,
where ∆G is the Gibbs free energy change, ∆H is the enthalpy change (in this case, -40 kJ), T is the temperature in Kelvin, and ∆S is the entropy change. For a reversible reaction at equilibrium, ∆G = 0.
At equilibrium, ∆G = 0, so we can write the equation as:
0 = -40 kJ - T∆S.
Rearranging the equation:
∆S = -(40 kJ / T).
Now, let's consider the activation energy (Ea) of both the forward and reverse reactions at equilibrium. At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. Mathematically, we can express this using the Arrhenius equation:
Rate forward = A * e^(-Ea forward / RT),
Rate reverse = A' * e^(-Ea reverse / RT),
where A and A' are the pre-exponential factors of the forward and reverse reactions, respectively, R is the gas constant, and T is the temperature in Kelvin.
Since the reaction is at equilibrium, the rates of the forward and reverse reactions are the same. Therefore, we can equate the two equations:
A * e^(-Ea forward / RT) = A' * e^(-Ea reverse / RT).
By dividing both sides of the equation by A' * e^(-Ea forward / RT), we get:
1 = (A / A') * e^((Ea forward - Ea reverse) / RT).
Taking the natural logarithm of both sides of the equation:
0 = ln(A / A') + ((Ea forward - Ea reverse) / RT).
As ∆G = 0, we have:
∆G = ∆H - T∆S = 0.
Substituting the value of ∆S = -(40 kJ / T), we get:
0 = -40 kJ - T * (-(40 kJ / T)) = -40 kJ + 40 kJ = 0.
Therefore, we can conclude that the activation energy of the reverse reaction (Ea reverse) is equal to the activation energy of the forward reaction (Ea forward). In this case, Ea reverse = Ea forward = 30 kJ.