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Consider the general reaction $$\mathrm{aA}+\mathrm{bB} \longrightarrow \mathrm{cC}$$ and the following average rate data over some time period \(\Delta t:\) $$-\frac{\Delta \mathrm{A}}{\Delta t}=0.0080 \mathrm{mol} / \mathrm{L} \cdot \mathrm{s}$$ $$-\frac{\Delta \mathrm{B}}{\Delta t}=0.0120 \mathrm{mol} / \mathrm{L} \cdot \mathrm{s}$$ $$\frac{\Delta \mathrm{C}}{\Delta t}=0.0160 \mathrm{mol} / \mathrm{L} \cdot \mathrm{s}$$ Determine a set of possible coefficients to balance this general reaction.

Short Answer

Expert verified
A possible set of coefficients to balance the general reaction is a = 2, b = 3, and c = 4, resulting in the balanced reaction: \(2\mathrm{A} + 3\mathrm{B} \longrightarrow 4\mathrm{C}\)

Step by step solution

01

Determine the relationship between the rates based on stoichiometry

The stoichiometry of the reaction states that, for every mole of A, b moles of B are consumed, resulting in the generation of c moles of C. So we can write the relationships as: \( \frac{Rate\_A}{a} = \frac{Rate\_B}{b} = \frac{Rate\_C}{c} \)
02

Use given average rate data

From the exercise, we have the average rate data: For A: \( -\frac{\Delta A}{\Delta t} = 0.0080\, \text{mol/L} \cdot \text{s}\\ \) For B: \( -\frac{\Delta B}{\Delta t} = 0.0120\, \text{mol/L} \cdot \text{s} \\ \) For C: \( \frac{\Delta C}{\Delta t} = 0.0160\, \text{mol/L} \cdot \text{s} \\ \)
03

Find the relationship between a and b

We have the relationship between the rates: \( \frac{Rate\_A}{a} = \frac{Rate\_B}{b} \Rightarrow \frac{0.0080}{a} = \frac{0.0120}{b} \Rightarrow \frac{a}{b} = \frac{0.0080}{0.0120} = \frac{2}{3} \) This tells us that the ratio of a to b is 2 : 3.
04

Find the relationship between b and c

We also have the relationship between the rates of B and C: \( \frac{Rate\_B}{b} = \frac{Rate\_C}{c} \Rightarrow \frac{0.0120}{b} = \frac{0.0160}{c} \Rightarrow \frac{b}{c} = \frac{0.0120}{0.0160} = \frac{3}{4} \) This tells us that the ratio of b to c is 3 : 4.
05

Find the coefficients a, b, and c

We know that \( \frac{a}{b} = \frac{2}{3} \) and \( \frac{b}{c} =\frac{3}{4} \). We can use these ratios to find a possible set of coefficients: Since the ratio of a to b is 2 : 3, we can use a = 2n and b = 3n, where n is a positive integer. Similarly, since the ratio of b to c is 3 : 4, we can use b = 3m and c = 4m, where m is a positive integer. Now, by substituting b in both equations, we get \( 3n = 3m \Rightarrow m=n \). To find the coefficients, we can choose n = 1, as choosing the smallest value helps in finding the simplest whole number coefficients. So, when n = 1, we have: a = 2, b = 3, c = 4. Therefore, a possible set of coefficients to balance the general reaction is a = 2, b = 3, and c = 4, resulting in the balanced reaction: \(2\mathrm{A} + 3\mathrm{B} \longrightarrow 4\mathrm{C}\)

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Key Concepts

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

Understanding Reaction Rate
The rate of a chemical reaction is a crucial concept in understanding how fast reactants transform into products. In terms of stoichiometry, the reaction rate is typically expressed in terms of the concentration of a reactant decreasing or the concentration of a product increasing over time.

For instance, given the average rate data provided in the original exercise, it's clear that reactants A and B are being consumed as they contribute to the formation of product C. By measuring the changes in concentration of A, B, and C over a given period, we can understand the dynamics of the reaction.

It's important to note that the reaction rate is dependent on various factors such as temperature, pressure, and the presence of a catalyst. These factors can significantly alter the speed at which a reaction occurs, which is essential when considering industrial applications or in a laboratory setting.
Chemical Kinetics Explained
Chemical kinetics involves the study of reaction rates and the way they change under various conditions, as well as the mechanisms that drive chemical reactions. The aim is to quantify the speed of a reaction and establish a relationship between the reaction conditions and the rate.

The exercise highlights a fundamental aspect of kinetics: the use of stoichiometry to relate different reaction rates. Kinetics intersects with stoichiometry when we use the stoichiometric coefficients to describe the proportional relationship between the rates of consumption and formation of reactants and products, respectively.

Rate Laws and Mechanisms

Kinetics also delves into the formulation of rate laws, which are mathematical expressions describing the relationship between the concentration of reactants and the reaction rate. Understanding the mechanism, or the step-by-step sequence of elementary reactions, allows scientists to propose the most likely pathway through which a reaction proceeds.
Balancing Chemical Equations
In any chemical reaction, it's essential to balance the equation to adhere to the law of conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction. By balancing chemical equations, we ensure that the same number of atoms of each element is present on both the reactant and product sides of the equation.

The process often involves finding the right stoichiometric coefficients, which are the numbers placed in front of compounds in a chemical equation. As demonstrated in the exercise, the coefficients must be adjusted so that the rate data for each substance aligns proportional to its stoichiometric coefficient. This ensures that the ratios reflect the actual number of molecules or moles of each substance that react, which is foundational in predicting product yields and in understanding the specifics of a chemical reaction.
Stoichiometric Coefficients Significance
Stoichiometric coefficients play a pivotal role in both balancing chemical equations and in stoichiometry, which is the quantitative relationship between reactants and products in a chemical reaction. These coefficients tell us in what ratio the molecules participate in the reaction.

In the provided example, by employing the ratios of the reaction rates and their corresponding stoichiometric coefficients, we were able to deduce a balanced chemical equation. The balanced equation is necessary for predicting how much product will form from given amounts of reactants and is fundamental in all aspects of chemistry, from the lab bench to industrial-scale chemical production.

Furthermore, these coefficients are indispensable for calculating the various aspects of a reaction, such as its yield, efficiency, and for scaling up from a lab experiment to commercial production, ensuring that the right quantities of materials are used and waste is minimized.

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Most popular questions from this chapter

A proposed mechanism for a reaction is $$\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Br} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{9}^{+}+\mathrm{Br}^{-} \quad \text { Slow }$$ $$\mathrm{C}_{4} \mathrm{H}_{9}^{+}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}_{2}^{+} \quad \text { Fast }$$ $$\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}_{2}^{+}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}+\mathrm{H}_{3} \mathrm{O}^{+}\quad \text { Fast }$$ Write the rate law expected for this mechanism. What is the overall balanced equation for the reaction? What are the intermediates in the proposed mechanism?

For enzyme-catalyzed reactions that follow the mechanism $$\mathbf{E}+\mathbf{S} \rightleftharpoons \mathbf{E} \cdot \mathbf{S}$$ $$\mathrm{E} \cdot \mathrm{S} \rightleftharpoons \mathrm{E}+\mathrm{P}$$ a graph of the rate as a function of \([\mathrm{S}]\), the concentration of the substrate, has the following appearance: Note that at higher substrate concentrations the rate no longer changes with [S]. Suggest a reason for this.

Which of the following statement(s) is(are) true? a. The half-life for a zero-order reaction increases as the reaction proceeds. b. A catalyst does not change the value of the rate constant. c. The half-life for a reaction, a \(A \longrightarrow\) products, that is first order in A increases with increasing \([\mathrm{A}]_{0}\) d. The half-life for a second-order reaction increases as the reaction proceeds.

The reaction $$\mathrm{NO}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)$$ was studied by performing two experiments. In the first experiment the rate of disappearance of NO was followed in the presence of a large excess of \(\mathrm{O}_{3}\). The results were as follows \(\left(\left[\mathrm{O}_{3}\right]\right.\) remains effectively constant at \(1.0 \times 10^{14}\) molecules/cm \(^{3}\) ): In the second experiment [NO] was held constant at \(2.0 \times 10^{14}\) molecules/cm \(^{3}\). The data for the disappearance of \(\mathbf{O}_{3}\) are as follows: a. What is the order with respect to each reactant? b. What is the overall rate law? c. What is the value of the rate constant from each set of experiments? $$\text { Rate }=k^{\prime}[\mathrm{NO}]^{x} \quad \text { Rate }=k^{\prime \prime}\left[\mathrm{O}_{3}\right]^{y}$$ d. What is the value of the rate constant for the overall rate law? $$\text { Rate }=k[\mathrm{NO}]^{\mathrm{x}}\left[\mathrm{O}_{3}\right]^y$$

A first-order reaction has rate constants of \(4.6 \times 10^{-2} \mathrm{s}^{-1}\) and \(8.1 \times 10^{-2} \mathrm{s}^{-1}\) at \(0^{\circ} \mathrm{C}\) and \(20 .^{\circ} \mathrm{C},\) respectively. What is the value of the activation energy?

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