Which of the following best explains the effects of a catalyst on the rate of a reversible reaction? |
It decreases the rate of the reverse reaction It increases the kinetic energy of the reacting molecules It moves the equilibrium position to the right It provides a new reaction path with a lower activation energy |
It provides a new reaction path with a lower activation energy |
The correct answer is option 4. It provides a new reaction path with a lower activation energy. To understand the effects of a catalyst on the rate of a reversible reaction, let us analyze each given option in detail: Catalysts are substances that increase the rate of a chemical reaction without being consumed in the process. They achieve this by providing an alternative reaction pathway with a lower activation energy. Evaluating the Options: 2. It increases the kinetic energy of the reacting molecules: This is also incorrect. A catalyst does not increase the kinetic energy of the reacting molecules. It lowers the activation energy barrier, allowing more molecules to have enough energy to react at a given temperature, but it doesn't change their inherent kinetic energy. 3. It moves the equilibrium position to the right: This is incorrect. A catalyst does not affect the position of equilibrium. It speeds up both the forward and reverse reactions equally, thus helping the system to reach equilibrium faster, but it does not change the equilibrium concentrations of reactants and products. 4. It provides a new reaction path with a lower activation energy: This is correct. A catalyst provides an alternative pathway for the reaction with a lower activation energy. This means that more reactant molecules have enough energy to overcome the activation energy barrier, thus increasing the rate of both the forward and reverse reactions. Conclusion: The best explanation for the effect of a catalyst on the rate of a reversible reaction is: It provides a new reaction path with a lower activation energy. This option correctly describes how catalysts work to increase the rate of both forward and reverse reactions by lowering the activation energy, thereby allowing the reaction to proceed more quickly to equilibrium without changing the equilibrium position itself. |