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A reaction was found to be second-order with respect to the concentration of carbon monoxide. If the concentration of carbon monoxide is doubled, with everything else kept the same, the rate of reaction will (a) remain unchanged (b) triple (c) increase by a factor of 4 (d) double

Short Answer

Expert verified
The rate of reaction will increase by a factor of 4. (Option c)

Step by step solution

01

Understand the Concept of Reaction Order

A second-order reaction means that the rate of the reaction depends on the square of the concentration of a reactant. In this case, the reaction is second-order with respect to carbon monoxide.
02

Identify the Rate Expression

For a second-order reaction with respect to carbon monoxide, the rate expression can be written as \( \text{Rate} = k[\text{CO}]^2 \), where \( k \) is the rate constant and \( [\text{CO}] \) is the concentration of carbon monoxide.
03

Determine the Effect of Doubling the Concentration

If the concentration of carbon monoxide \([\text{CO}])\) is doubled, the new concentration \([\text{CO}]_{new}\) will be \(2[\text{CO}]\). Substitute this into the rate expression: \( \text{Rate}_{new} = k(2[\text{CO}])^2 = 4k[\text{CO}]^2 \).
04

Compare the New Rate to the Initial Rate

The original rate expression was \( \text{Rate} = k[\text{CO}]^2 \). The new rate expression \( \text{Rate}_{new} = 4k[\text{CO}]^2 \) shows that the new rate is four times the original rate.
05

Conclusion

Because doubling the concentration of carbon monoxide results in the rate being multiplied by 4, the correct answer is option (c): increase by a factor of 4.

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Key Concepts

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

Understanding Reaction Order
In chemical kinetics, the reaction order is a fundamental concept that tells us how the rate of a chemical reaction depends on the concentration of the reactants. It is determined by the sum of the powers to which all reactant concentrations are raised in the rate expression. For example, if a reaction is said to be second-order with respect to carbon monoxide (CO), this means that the rate of the reaction is proportional to the square of the concentration of CO. This way, reaction order helps us predict how changes in reactant concentrations will affect the overall rate of the reaction. It's important to note that reaction order is not necessarily the same as the stoichiometric coefficients in a balanced chemical equation.
Decoding the Rate Expression
The rate expression, or rate law, is an equation that links the rate of a reaction to the concentration of reactants, often incorporating a rate constant. For a second-order reaction with respect to a single reactant such as carbon monoxide, the rate expression can be written as:\[ \text{Rate} = k[\text{CO}]^2 \]In this formula, \( k \) is the rate constant, a unique constant value for each chemical reaction under specific conditions. The rate expression shows how the reaction rate is related to the concentration of reactants raised to the power that reflects their order in the reaction. A second-order means the concentration factor is squared, meaning any change in the concentration of CO will have an exponential effect on the reaction rate.
Analyzing Reaction Rate Changes
Reaction rate is essentially a measure of how quickly or slowly reactants are converted into products in a chemical reaction. It is crucial to understand how changes in concentration affect the reaction rate, especially when dealing with second-order kinetics. In a second-order reaction, because the concentration of CO is squared in the rate expression, doubling the concentration of CO will lead to a fourfold increase in the reaction rate. This is calculated by substituting the new concentration into the rate expression. When you plug \( 2[\text{CO}] \) into \( (2[\text{CO}])^2 = 4[\text{CO}]^2 \), you see that the rate becomes four times greater, illustrating the power of reaction order on rate change.
Exploring Carbon Monoxide Concentration Impact
Carbon monoxide concentration plays a pivotal role in determining the reaction rate for the process under consideration. In reactions where CO acts as a reactant, even small adjustments in its concentration can lead to significant changes in how fast the reaction proceeds. For example, increasing the concentration of CO results in more molecules available to collide and react, leading to an increased rate. Specifically for a second-order reaction, doubling the concentration results in a rate that multiplies by four. Therefore, understanding and controlling the concentration of CO is crucial in industrial and chemical processes where reaction efficiency and speed are important.

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Most popular questions from this chapter

Which of the following statements is correct? (1) order of a reaction can be known from experimental results and not from the stoichiometry of a reaction. (2) molecularity a reaction refers to (i) each of the elementary steps in (an overall mechanism of) a complex reaction or (ii) a single step reaction. (3) overall molecularity of a reaction may be determined in a manner similar to overall order of reaction. (4) overall order of a reaction \(\mathrm{A}^{\mathrm{m}}+\mathrm{B}^{\mathrm{n}} \longrightarrow \mathrm{AB}_{\mathrm{x}}\) is \(\mathrm{m}+\mathrm{n}\) Select the correct answer using the following codes: (a) 2 and 3 (b) 1,3 and 4 (c) 2,3 and 4 (d) 1,2 and 3

If the rate law of a reaction \(\mathrm{nA} \longrightarrow \mathrm{B}\) is expressed as Rate \(=-\frac{1}{n} \frac{d[A]}{d t}=+\frac{d[B]}{d t}=k[A]^{x}\) The unit of the rate constant will be (a) \(\mathrm{mol}^{\mathrm{x}} / \mathrm{L}^{\mathrm{x}} / \mathrm{s}\) (b) \(\mathrm{L}^{\mathrm{x}} / \mathrm{mol}^{\mathrm{t}} \mathrm{s}\) (c) \(m o l^{(1-x)} / L^{(x-1)} \cdot S^{-1}\) (d) \(\operatorname{mol}^{(x-1)} / L^{(1-x)} \cdot S^{-1}\)

The rate constant of first-order reaction is \(3 \times 10^{-6}\) per second. The initial concentration is \(0.10 \mathrm{M}\). The initial rate is (a) \(3 \times 10^{-7} \mathrm{Ms}^{-1}\) (b) \(3 \times 10^{-4} \mathrm{Ms}^{-1}\) (c) \(3 \times 10^{-5} \mathrm{Ms}^{-1}\) (d) \(3 \times 10^{-6} \mathrm{Ms}^{-1}\)

The rate constant for the reaction, \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\) is \(3.0 \times 10^{-5} \mathrm{~s}^{1} .\) If the rate is \(2.40 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{1}\), then the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) (in \(\mathrm{mol} \mathrm{L}^{4}\) ) is (a) \(1.4\) (b) \(1.2\) (c) \(0.04\) (d) \(0.8\)

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

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