Question

The energies of activation for forward and reverse reactions for $({{A}_{2}}+{{B}_{2}}\rightleftharpoons 2AB)$ are $180kJ\,mo{{l}^{-1}}$ and $200kJ\,mo{{l}^{-1}}$ respectively. The presence of a catalyst lowers the activation energy of both (forward and reverse) reactions by $100kJ\,mo{{l}^{-1}}$ . The enthalpy change of the reaction $({{A}_{2}}+{{B}_{2}}\to 2AB)$ in the presence of catalyst will be (in $kJ\,mo{{l}^{-1}}$)
A) $300$
B) $120$
C) $280$
D) $20$

Answer 1

To increase the rate of reaction, catalyst is added without getting consumed in the process. It helps in speeding up the reaction, by decreasing the activation energy or changing the reaction mechanism. Enzymes act as catalysts in biochemical reactions.

Complete step by step solution:
In the absence of catalyst energy, the activation energy for forward reaction is $180kJ\,mo{{l}^{-1}}$
In the absence of catalyst energy, the activation energy for forward reaction is $200kJ\,mo{{l}^{-1}}$
Catalyst lowers the activation energy of both the reactions by $100kJ\,mo{{l}^{-1}}$
Hence, in the presence of catalyst, the activation energy for forward energy will decrease by $100kJ\,mo{{l}^{-1}}$
That is, $180-100$
$\Rightarrow 80kJ\,mo{{l}^{-1}}$
In the presence of catalyst, the activation energy for forward energy will decrease by $100kJ\,mo{{l}^{-1}}$
That is, $200-100$
$\Rightarrow 100kJ\,mo{{l}^{-1}}$
To calculate enthalpy change, a formula is used:
$\Delta {{H}_{R}}={{E}_{f}}-{{E}_{b}}$
where, $\Delta {{H}_{R}}$ is the enthalpy change of the reaction, ${{E}_{f}}$ is the activation energy for forward reaction and ${{E}_{b}}$ is the activation energy for backward reaction.
Now, substituting the values in the above formula, we get
$\Delta {{H}_{R}}=80-100$
$\Rightarrow -20kJ\,mo{{l}^{-1}}$

Hence, the correct option is (D).

Note: Activation energy does not depend upon the temperature, concentration, pressure and volume.Negative catalysts are also known as inhibitors, which slow down the rate of reaction.When a reactant molecule collides with each other at the highest energy point, then it will result in the formation of an intermediate, which remains in equilibrium with the reactant. If the energy of this intermediate complex is equal to or greater than the threshold energy, then it converts into products.