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Chapter 12: Problem 31

In a reversible reaction, a catalyst (a) increases the rate of forward reaction only (b) increases the rate of forward reaction to a greater extent that of the backward reaction (c) increases the rate of forward reaction and decreases that of the backward reaction (d) increases the rate of forward and backward reaction by the same factor.

Short Answer

Expert verified
(d) increases the rate of forward and backward reaction by the same factor.

Step by step solution

01

Understanding Reversible Reactions

Firstly, understand that in a reversible reaction, the reactants can form products, and simultaneously, the products can revert back to reactants. This occurs in a state of dynamic equilibrium.
02

Role of a Catalyst

A catalyst is a substance that increases the rate of a chemical reaction without undergoing any permanent chemical change itself. It does this by providing a new pathway with a lower activation energy for the reaction, which is equally viable for both the forward and reverse reactions.
03

Effect of Catalyst on Reversible Reactions

Since a catalyst offers a lower activation energy pathway, both the forward and the reverse reactions are sped up. Importantly, this acceleration does not affect the equilibrium state of the reaction; it simply helps the system reach equilibrium more quickly.
04

Identifying the Correct Option

Given the role of a catalyst in reversible reactions, we can conclude that the correct answer is the one that states a catalyst increases the rates of both the forward and backward reactions equally.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equilibrium
Now, when students approach problems involving equilibrium, it's crucial to consider not just the state at which the reaction sits but how it arrived there and how changes will affect it. For instance, increasing the temperature may favor one side of the equilibrium鈥攁nd understanding this can help students predict and control chemical reactions effectively.
Activation Energy
It is vital to remember, though, that activation energy is just that鈥攁n initial hurdle. Lowering it does not alter the final balance of reactants and products, nor does it influence how much energy is released or absorbed in the reaction overall. It simply makes reaching equilibrium faster.
Rate of Chemical Reactions
A catalyst is often used to increase the rate of a reaction, making it a central point of interest, especially in industrial processes where time is money. By lowering the activation energy, as discussed earlier, a catalyst permits more particles to react over the same period, thus speeding up both the forward and backward reactions in a reversible process. Nonetheless, despite hastening the journey to equilibrium, the catalyst does not change the position of the equilibrium or the proportions of reactants and products at equilibrium, ensuring that the chemistry remains unchanged even as the tempo changes.

Understanding these principles allows students to manipulate reaction conditions for desired outcomes in chemical processes, be it in a laboratory or commercial setting.

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Most popular questions from this chapter

Which of the following statements is false for catalyst? (a) A catalyst initiates the reaction. (b) It does not alter the position of equilibrium in a reversible reaction. (c) A catalyst remains unchanged in quantity and composition at the end of reaction. (d) Catalysts are sometimes very specific in respect of a reaction.

The following information is available for the reaction: \(\mathrm{R} \rightarrow \mathrm{P}\) : \(\mathrm{R} \longrightarrow \mathrm{P} ; \quad\) rate \(=r_{1}\) \(\mathrm{R} \stackrel{\mathrm{A}}{\longrightarrow} \mathrm{P} ; \quad\) rate \(=r_{2}\) \(\mathrm{R} \stackrel{\mathrm{B}}{\longrightarrow} \mathrm{P} ; \quad\) rate \(=r_{3}\) \(\mathrm{R} \stackrel{\mathrm{c}}{\longrightarrow} \mathrm{P} ; \quad\) rate \(=r_{4}\) \(\mathrm{R} \stackrel{\mathrm{D}}{\longrightarrow} \mathrm{P} ; \quad\) rate \(=r_{5}\) \(\mathrm{R} \stackrel{\mathrm{A}+\mathrm{C}}{\longrightarrow} \mathrm{P} ; \quad\) rate \(=r_{6}\) \(\mathrm{R} \stackrel{\mathrm{A}+\mathrm{D}}{\longrightarrow} \mathrm{P} ; \quad\) rate \(=r_{7}\) If \(r_{3}

Tanning of leather is (a) colouring of leather by chemicals. (b) drying process to make the leather hard. (c) polishing of leather to make it look attractive. (d) coagulative hardening of the leather by chemicals.

The correct match is $$ \begin{array}{ll} \hline \text { Column I } & \text { Column II } \\ \hline \text { (1) Coagulation } & \text { (P) } \begin{array}{l} \text { Scattering of } \\ \text { light } \end{array} \\ \text { (2) Peptization } & \text { (Q) } \begin{array}{l} \text { Purification } \\ \text { of colloidal } \\ \text { solution } \end{array} \\ \begin{array}{ll} \text { (3) } \text { Tyndall } \\ \text { effect } \end{array} & \text { (R) } \begin{array}{l} \text { Addition of an } \\ \text { electrolyte } \end{array} \\ \text { (4) Dialysis } & \text { (S) } \begin{array}{l} \text { Precipitation } \\ \text { of colloidal } \\ \text { solution } \end{array} \\ \hline \end{array} $$ $$\begin{array}{llll}1 & 2 & 3 & 4\end{array}$$ (a) \(\begin{array}{llll}\mathrm{P} & \mathrm{Q} & \mathrm{R} & \mathrm{S}\end{array}\) (b) \(\begin{array}{llll}\mathrm{S} & \mathrm{R} & \mathrm{P} & \mathrm{Q}\end{array}\) (c) \(\begin{array}{llll}\mathrm{R} & \mathrm{S} & \mathrm{Q} & \mathrm{P}\end{array}\) (d) Q \(\mathrm{R} \quad \mathrm{P} \quad \mathrm{S}\)

A quantity of \(1.9 \times 10^{-4} \mathrm{~g}\) of the metal having density \(19 \mathrm{~g} / \mathrm{ml}\) is dispersed in 11 of water to give a sol having spherical metal particles of radius \(10 \mathrm{~nm} .\) The approximate number of metal sol particles per \(\mathrm{cm}^{3}\) of the sol is (a) \(2.35 \times 10^{9}\) (b) \(4 \times 10^{10}\) (c) \(1.9 \times 10^{9}\) (d) \(2.8 \times 10^{9}\)
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