For each of the questions, four choices have been provided. Select the correct alternative. The equilibrium constant \(\mathrm{K}_{\mathrm{c}}\) is \(10^{2}\) for the reaction \(\mathrm{AB}+\mathrm{C} \rightleftarrows \mathrm{AC}+\mathrm{B}\) The rate constant for the forward reaction \(\mathbf{K}\) is 106, the rate constant of backward reaction is (1) \(10^{4}\) (2) \(10^{8}\) (3) \(10^{-4 \mathrm{v}}\) (4) \(\frac{1}{100}\)

Short Answer

Expert verified
Answer: The rate constant of the backward reaction is \(10^4\).

Step by step solution

01

Write down the given information and the relationship between equilibrium constant and rate constants

We are given: - Equilibrium constant, \(\mathrm{K_c} = 10^2\) - Rate constant of the forward reaction, \(\mathrm{k} = 10^6\) The relationship between equilibrium constant \(\mathrm{K_c}\) and rate constants of the forward and backward reactions is given by: $$ \mathrm{K_c} = \frac{k}{k'},$$ where - \(k\) is the rate constant of the forward reaction - \(k'\) is the rate constant of the backward reaction Our goal is to find the rate constant of the backward reaction, \(k'\).
02

Solve for the rate constant of the backward reaction, \(k'\)

Using the relationship we obtained in step 1, we can solve for \(k'\) as follows: $$ k' = \frac{k}{\mathrm{K_c}}$$ Plugging the given values: $$ k' = \frac{10^6}{10^2}$$ $$ k' = 10^4$$
03

Compare the calculated value of \(k'\) with the given options

Our calculated value for the rate constant of the backward reaction, \(k'\), is \(10^4\). Comparing this with the given options, we find that it matches option (1). Thus, the correct answer is option (1) \(10^4\).

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Key Concepts

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

Chemical Kinetics
Chemical kinetics is the study of the rates at which chemical reactions proceed and the factors that influence these rates. In educational terms, think of it as the 'who,' 'what,' 'when,' and 'how' of a chemical reaction. The 'who' comprises the reactants and products, the 'what' refers to the actual change from reactants to products, the 'when' is the reaction rate, and the 'how' involves the path or mechanism through which the reaction occurs.

Understanding kinetics can answer questions like how long a reaction will take to reach completion or what conditions are needed to achieve a certain product yield. It's not just about the speed; factors such as temperature, concentration, and catalyst presence also play crucial roles in how quickly reactions occur.
Rate Constant
The rate constant is a crucial factor in chemical kinetics. It's like the 'pace-setter' of a chemical reaction. Represented by the symbol 'k' for a forward reaction and 'k' prime for a backward reaction, it is unique to each reaction and its specific conditions, like a fingerprint. High school or introductory college courses typically explain that the rate constant is influenced by temperature and the presence of a catalyst, which acts like a personal trainer for chemical reactions – speeding them up without participating directly.

A larger rate constant indicates a faster reaction. When students are solving problems involving rate constants, they use them to calculate reaction rates or to find equilibrium constants. It's a measure of the propensity of a reaction to move towards products, and it's typically determined experimentally.
Reaction Rate
The reaction rate is essentially the velocity of a chemical reaction. It measures how quickly reactants get converted into products over time. It's like watching a sped-up video of a flower blooming - the faster it blooms, the higher the reaction rate. This rate can be expressed in terms of the decrease in concentration of reactants or the increase in concentration of products per unit time.

In educational exercises, students often calculate the reaction rate using the rate law, which links the reaction rate to the concentrations of reactants and the rate constant. Reaction rates are integral to understanding chemical processes, such as those in industry, environmental chemistry, and even cooking!
Dynamic Equilibrium
Picture a crowded dance floor where for every person that exits, another enters - that's the concept of dynamic equilibrium in a chemical reaction. It's a state in which the rate of the forward reaction equals the rate of the backward reaction, so there's no net change in the amounts of reactants and products.

It's 'dynamic' because the reactions haven't stopped; both the forward and reverse reactions are still happening. Yet, it's in 'equilibrium' because everything is in balance - the concentrations of reactants and products remain constant over time. The equilibrium constant (K) is the numerical value that represents the ratio of product concentrations to reactant concentrations at equilibrium, with each raised to the power of their stoichiometric coefficients. In simpler terms, it's a number that tells you the 'happy medium' where all the chemical dancers are content to maintain their current state of back-and-forth.

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Most popular questions from this chapter

For each of the questions, four choices have been provided. Select the correct alternative. Which of the following is true? (a) In an endothermic equilibrium reaction, activation energy required for a forward reaction is higher than that for a backward reaction. (b) To an endothermic equilibrium reaction, activation energy required to forward reaction is lower than that for a backward reaction. (c) Activation energy required for both forward and backward reactions is same in equilibrium. (d) No activation energy is required for an exothermic backward equilibrium reaction.

For each of the questions, four choices have been provided. Select the correct alternative. In the formation of \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) from \(\mathrm{NO}_{2}\) the rates of production of (a) \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) are equal (b) \(\mathrm{NO}\) is double the rate of consumption of \(\mathrm{NO}_{2}\) (c) \(\mathrm{NO}\) is twice the rate of production of \(\mathrm{O}_{2}\) (d) \(\mathrm{O}_{2}\) is twice the rate of production of \(\mathrm{NO}\)

The minimum energy which two molecules should possess so that their collisions result in a chemical reaction is called_________ energy.

For each of the questions, four choices have been provided. Select the correct alternative. Equilibrium position of which of the following reactions is not affected by change in pressure? (a) \(\mathrm{I}_{2(\mathrm{~s})}+5 \mathrm{~F}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{IF}_{5(\mathrm{~g})}\) (b) \(\mathrm{FeO}_{(\mathrm{s})}+\mathrm{CO}_{(\mathrm{g})} \rightarrow \mathrm{Fe}_{(\mathrm{s})}+\mathrm{CO}_{(\mathrm{g})}\) (c) \(2 \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2(\mathrm{~s})} \rightarrow 2 \mathrm{CuO}_{(\mathrm{s})}+4 \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}\) (d) \(\mathrm{N}_{2} \mathrm{O}_{4(\mathrm{~g})} \rightarrow 2 \mathrm{NO}_{2(\mathrm{~g})}\)

\(2 \mathrm{SO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \rightleftarrows 2 \mathrm{SO}_{3(\mathrm{~g})}+\mathrm{Q} \mathrm{kJ}\) In the above reaction, how can the yield of product be increased without increasing the pressure? (a) by increasing temperature (b) by decreasing temperature (c) by increasing the volume of the reaction vessel (d) by the addition of the catalyst

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