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The reaction between chlorine and nitric oxide to form nitrosyl chloride is shown below. If the reaction rate doubles when the concentration of \(\mathrm{Cl}_{2}\) is doubled and the rate quadruples when the concentration of \(\mathrm{NO}\) is doubled, by what factor will the rate increase if both concentrations, \(\mathrm{NO}\) and \(\mathrm{Cl}_{2}\) are doubled? \(\mathrm{Cl}_{2}(\mathrm{~g})+2 \mathrm{NO}(\mathrm{g}) \rightarrow 2 \mathrm{NOCl}(\mathrm{g})\) a. 4 b. 8 c. 12 d. 16

Short Answer

Expert verified
The rate will increase by a factor of 8.

Step by step solution

01

Identify the Rate Law

From the information given, we know: doubling the concentration of \( \mathrm{Cl}_2 \) doubles the reaction rate and doubling the concentration of \( \mathrm{NO} \) quadruples the reaction rate. This suggests the rate law is \( \text{Rate} = k[\mathrm{Cl}_2]^1[\mathrm{NO}]^2 \), where "\( k \)" is the rate constant.
02

Analyze Effect of Doubling Concentrations

According to the rate law \( \text{Rate} = k[\mathrm{Cl}_2]^1[\mathrm{NO}]^2 \), if we double \( [\mathrm{Cl}_2] \), the multiplier is \( 2^1 = 2 \), and if we double \( [\mathrm{NO}] \), the multiplier is \( 2^2 = 4 \).
03

Determine Combined Effect

If both \( [\mathrm{Cl}_2] \) and \( [\mathrm{NO}] \) are doubled simultaneously, we multiply their individual effects: \( 2 \times 4 = 8 \).
04

Conclusion

Therefore, doubling both \( [\mathrm{Cl}_2] \) and \( [\mathrm{NO}] \) will increase the reaction rate by a factor of 8.

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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 occur. It helps us understand how different factors, like temperature, concentration, and catalysts, affect the speed of a reaction. Understanding kinetics is crucial because it allows chemists to control reactions, either speeding them up or slowing them down depending on the need.

In the context of chemical kinetics, the reaction between chlorine (\(\mathrm{Cl}_2\)) and nitric oxide (\(\mathrm{NO}\)) that forms nitrosyl chloride (\(\mathrm{NOCl}\)) is a prime example. This reaction happens when molecules collide with enough energy to form a product. In our scenario, determining the reaction rate involves examining how the concentrations of the reactants affect the rate. The speed of the reaction can tell us how quickly the products form, which is vital for both laboratory and industrial chemical processes.

Chemical kinetics doesn’t just describe the speed of a reaction, it also helps predict the concentrations of reactants and products at any given time during the reaction. This insight is essential not only for chemists but also for engineers and scientists involved in designing processes and reactors.
Rate Law Expression
Rate law expressions are mathematical formulas that show how the speed of a chemical reaction depends on the concentration of its reactants. Unlike the balanced equation, which tells us the stoichiometry of the reaction, the rate law must be determined experimentally.

For the reaction \(\mathrm{Cl}_2 + 2\mathrm{NO} \rightarrow 2\mathrm{NOCl}\), the rate law is given as \(\text{Rate} = k[\mathrm{Cl}_2]^1[\mathrm{NO}]^2\). This tells us that the reaction rate is directly proportional to the concentration of chlorine and the square of the concentration of nitric oxide. Here, "\(k\)" represents the rate constant, a unique value for each reaction at a given temperature. It remains unchanged unless the conditions of the reaction, like temperature, change.

In this particular reaction, the order with respect to \(\mathrm{Cl}_2\) is 1, and with respect to \(\mathrm{NO}\) is 2. Therefore, if the concentration of \(\mathrm{Cl}_2\) is doubled, the rate doubles. If \(\mathrm{NO}\)'s concentration is doubled, the rate increases by a factor of four, owing to the square term. Understanding the rate law is crucial for predicting how changes in concentration affect the reaction rate.
Effect of Concentration on Reaction Rate
The effect of concentration on reaction rate is a fundamental concept in chemical kinetics. When concentrations of reactants change, the rate of the reaction typically changes as well. In general, increasing the concentration of reactants increases the rate of reaction because there are more molecules available to collide and react.

In our specific example, increasing the concentration of \(\mathrm{Cl}_2\) and \(\mathrm{NO}\) leads to more frequent collisions between the molecules. These more frequent collisions result in a faster formation of \(\mathrm{NOCl}\). The exercise illustrates this: doubling the concentration of \(\mathrm{Cl}_2\) doubles the rate, while doubling \(\mathrm{NO}\) quadruples it. Both effects combined, when \(\mathrm{Cl}_2\) and \(\mathrm{NO}\) concentrations are doubled, lead to an eightfold increase in the rate of the reaction.

This relationship between concentration and rate is critical in many practical scenarios, such as industrial chemical production, where maximizing reaction rates can lead to more efficient processes and higher yields.

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Most popular questions from this chapter

Which of the following statement about the Arrhenius equation is/are correct? a. On raising temperature, rate constant of the reaction of greater activation energy increases less rapidly than that of the reaction of smaller activation energy. b. The term \(\mathrm{e}^{-E a / \mathrm{RT}}\) represents the fraction of the molecules having energy in excess of threshold value. c. The pre-exponential factor becomes equal to the rate constant of the reaction at extremely high temperature. d. When the activation energy of the reaction is zero, the rate becomes independent of temperature

Two reactions \(\mathrm{X} \rightarrow\) Products and \(\mathrm{Y} \rightarrow\) products have rate constant \(\mathrm{k}_{\mathrm{x}}\) and \(\mathrm{k}_{\mathrm{Y}}\) at temperature \(\mathrm{T}\) and activation energies \(\mathrm{E}_{\mathrm{x}}\) and \(\mathrm{E}_{\mathrm{Y}}\) respectively. If \(\mathrm{k}_{\mathrm{x}}>\) \(\mathrm{k}_{\mathrm{r}}\) and \(\mathrm{E}_{\mathrm{x}}<\mathrm{E}_{\mathrm{Y}}\) and assuming that for both the reaction is same, then a. At lower temperature \(\mathrm{k}_{\mathrm{Y}}>\mathrm{k}_{\mathrm{x}}\) b. At higher temperature \(\mathrm{k}_{\mathrm{x}}\) will be greater than \(\mathrm{k}_{\mathrm{y}}\) c. At lower temperature \(\mathrm{k}_{\mathrm{x}}\) and \(\mathrm{k}_{\mathrm{Y}}\) will be close to each other in magnitude d. At temperature rises, \(\mathrm{k}_{\mathrm{x}}\) and \(\mathrm{k}_{\mathrm{Y}}\) will be close to each other in magnitude

When the reactants are \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) at one mole per litre each the rate equation is, rate \(=\mathrm{k}[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{1 / \mathrm{Y}}\) \([\mathrm{C}]^{\mathrm{X} / \mathrm{Y}}\). The order of the reaction is a. \(X+\frac{(1+X)}{Y}\) b. \(\mathrm{X}-\mathrm{Y}+\frac{\mathrm{X}}{\mathrm{Y}}\) c. \(\mathrm{X}+\mathrm{Y}+\frac{\mathrm{X}}{\mathrm{Y}}\) d. \(2(X+Y)\)

Consider a reaction \(\mathrm{aG}+\mathrm{bH} \rightarrow\) Products. When concentration of both the reactants \(\mathrm{G}\) and \(\mathrm{H}\) is doubled, the rate increases by eight times. However when concentration of \(\mathrm{G}\) is doubled keeping the concentration of \(\mathrm{H}\) fixed, the rate is doubled. The overall order of the reaction is a. 0 b. 1 c. 2 d. 3

The rate law has the form; Rate \(=\mathrm{k}[\mathrm{A}][\mathrm{B}]^{3 / 2}\), can the reaction be an elementary process? a. yes b. no c. may be yes or no d. can not be predicted

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