Question

(2 points) A reaction has a forward activation energy of 52 kJ and a reverse activation energy of 19 kJ. 5. a. What is the AH of the forward reaction? b. Is the forward reaction endothermic or exothermic? What happens to the activation energy of the forward reaction if the temperature is increased? c. d. At the same tempraue and equal concentrations of reactatswill seod reaction with a forward activation energy of 100 kJ be faster or slower than this reaction? Why?

Answer 1

Explanation:

(a)  It is given that value of forward activation energy is 52 kJ and the value of reverse activation energy is 19 kJ.

Hence, calculate the value of $\Delta H$ as follows.

     $\Delta H = E_{a}_{reverse} - E_{a}_{forward}$

                    = 19 kJ - 52 kJ

                    = -33 kJ

(b)   As it is known that when value of $\Delta H$ is negative then it means heat is releasing and the reaction is exothermic in nature.

And, when value of $\Delta H$ is positive then it means heat is absorbed by the reactant molecules and the reaction is endothermic in nature.

Since, the value of $\Delta H$ for given reaction is negative. Hence, it is an exothermic reaction.

(c)   The minimum amount of energy necessary to carry out a chemical reaction is known as activation energy.

So, when we increase the temperature then there will also occur an increase in number of collisions between the particles. But the energy required to start the reaction will remain the same.

Therefore, if the temperature is increased then activation energy will remain the same.

(d)    When higher is the value of activation energy then it means higher is the energy barrier. Hence, it becomes difficult for reactant molecules to cross that energy barrier.

So, when value of $E_{a}_{forward}$ is 100 kJ then reaction will become slow.