Solveeit Logo
Login

Question

Question: In an exothermic reaction \(X\to Y\), the activation energy is \(100\text{ kJmo}{{\text{l}}^{-1}}\) ...

In an exothermic reaction X→YX\to Y, the activation energy is 100 kJmol−1100\text{ kJmo}{{\text{l}}^{-1}} of X. The enthalpy of the reaction is −140 kJmol−1-140\text{ kJmo}{{\text{l}}^{-1}}. The activation energy of the reverse reaction Y→XY\to X is:
A. 40 kJmol−140\text{ kJmo}{{\text{l}}^{-1}}
B. 340 kJmol−1340\text{ kJmo}{{\text{l}}^{-1}}
C. 240 kJmol−1240\text{ kJmo}{{\text{l}}^{-1}}
D. 100 kJmol−1100\text{ kJmo}{{\text{l}}^{-1}}

Explanation

Solution

Activation energy for the forward reaction is already given and the enthalpy for the reaction is also given. We have to find out the activation energy of the backward reaction. Remember, the enthalpy of the reaction equals to the difference between forward and the backward reaction.

Complete step by step solution:
Given that, For the exothermic reaction, the activation energy (consider Ef{{E}_{f}}) is 100 kJmol−1100\text{ kJmo}{{\text{l}}^{-1}}.
The enthalpy for the reaction is −140 kJmol−1-140\text{ kJmo}{{\text{l}}^{-1}}.
We have to find out the activation energy for the reverse reaction (consider Er{{E}_{r}}).
So, let us first know what is activation energy and enthalpy. Activation energy is the extra amount of energy that is given to get useful work done. In chemistry, the activation energy is generally called as the minimum amount of energy (or threshold energy) needed to activate or to energize the molecules or atoms to undergo a chemical reaction. While, electron gain enthalpy is the amount of energy which gets released upon the acceptance of an atom from any neutral atom to form negative gaseous ions in a process. The enthalpy of a reaction can be calculated by differentiating the activation energy required for forward and reverse reaction. So, let’s consider enthalpy as ΔH\Delta H which is given as −140 kJmol−1-140\text{ kJmo}{{\text{l}}^{-1}} and the formula is written as:
ΔH=Ef−Er\Delta H={{E}_{f}}-{{E}_{r}}
Thus, Er=Ef−ΔH{{E}_{r}}={{E}_{f}}-\Delta H
Now, by placing the values we get:
Er=100−(−140)=240 kJmol−1{{E}_{r}}=100-(-140)=240\text{ kJmo}{{\text{l}}^{-1}}

Hence, the correct option is C.

Note: It is important to keep in mind, if the enthalpy is having a negative magnitude it means the reaction is an exothermic reaction i.e. it releases energy or heat during the reaction while in an endothermic reaction, heat or energy is consumed.