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Chapter 14: Problem 2

Distinguish between average rate and instantaneous rate. Which of the two rates gives us an unambiguous measurement of reaction rate? Why?

Short Answer

Expert verified
The instantaneous rate gives an unambiguous reaction rate measurement because it is specific to a moment in time.

Step by step solution

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01

Understanding Rates

Begin by understanding the two types of rates: average rate and instantaneous rate. The average rate is calculated over a finite period of time, while the instantaneous rate is the rate at a specific moment.
02

Calculating Average Rate

The average rate of a reaction can be determined by taking the change in concentration of a reactant or product divided by the time interval over which the change occurs. Mathematically it is given as: \[ \text{Average Rate} = \frac{\Delta [ ext{Concentration}]}{\Delta t} \] where \( \Delta [\text{Concentration}] \) is the change in concentration and \( \Delta t \) is the change in time.
03

Calculating Instantaneous Rate

To determine the instantaneous rate, one needs to find the derivative of the concentration with respect to time at a specific point. This involves calculus and is represented as: \[ \text{Instantaneous Rate} = \frac{d[ ext{Concentration}]}{dt} \]
04

Comparing Rates

When comparing the two, the instantaneous rate gives a specific and precise measurement of the reaction rate at a particular time point, whereas the average rate may obscure fluctuations by smoothing them over a time period.
05

Determining Unambiguous Measurement

The instantaneous rate gives an unambiguous measure of the reaction rate because it reflects the exact rate at a specific time without averaging over a time period, capturing rapid changes and precise behavior of the reaction.

Key Concepts

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

Average Rate
The average rate of a reaction provides a broad view of how fast a reaction is occurring over a certain period.
  • It is calculated by taking the change in concentration of a substance and dividing it by the time interval during which this change occurs.
  • Mathematically, it can be expressed as: \( \text{Average Rate} = \frac{\Delta [\text{Concentration}]}{\Delta t} \), where \( \Delta [\text{Concentration}] \) is the change in concentration and \( \Delta t \) represents the time interval.

A helpful analogy is thinking about a car trip. The average speed can be determined by dividing the total distance by the total time taken. Similarly, average rate tells us how overall change happens during the reaction, but may not reflect any fluctuations within the time span.
Instantaneous Rate
Instantaneous rate provides a precise measurement of how fast a reaction proceeds at a particular moment.
  • To find the instantaneous rate, you focus on a single point in time, much like checking a car's speedometer to find out how fast it's going at that very second.
  • This measurement can be found using calculus, particularly through the derivative of concentration with respect to time: \( \text{Instantaneous Rate} = \frac{d[\text{Concentration}]}{dt} \).

This method allows for capturing the most current reaction speed, reflecting immediate changes and offering a clear picture of the reaction's behavior at a specific point.
Derivative in Calculus
The derivative in calculus is a powerful tool applied in determining the instantaneous rate of reaction. It measures how a function changes at any given point and is essential for finding rates.
  • The rate at which concentration changes over a small time period can be described by the derivative \( \frac{d[\text{Concentration}]}{dt} \).
  • This mathematical concept is used to find the slope of the tangent line to the concentration vs. time graph at a specific point.

The derivative helps in pinpointing the exact rate of reaction at any given second, aiding in analyzing rapid changes and intricate details that average rate might miss.
Concentration Change
Concentration change is a fundamental aspect in calculating both average and instantaneous rates of reaction.
  • Changes in concentration of either reactants or products help determine the rate at which chemical processes occur.
  • In the context of rates, \( \Delta [\text{Concentration}] \) represents the difference in initial and final concentrations over a time period \( \Delta t \).

Understanding how concentration changes over time offers insight into the progress and speed of a reaction, allowing us to measure both short-term (instantaneous) and long-term (average) changes.
Time Interval in Reactions
The time interval in reactions is crucial for calculating an accurate reaction rate, especially for determining average rates.
  • It is symbolized by \( \Delta t \), representing the period over which a reaction’s progress is measured.
  • Smaller time intervals offer a clearer picture of fluctuations in reaction speed, especially when determining instantaneous rate.

Choosing the correct time interval is essential, as longer intervals might smooth out rapid changes, yielding average rates that may not reflect specific rate dynamics occurring at particular moments.

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Most popular questions from this chapter

To carry out metabolism, oxygen is taken up by hemoglobin \((\mathrm{Hb})\) to form oxyhemoglobin \(\left(\mathrm{Hb} \mathrm{O}_{2}\right)\) according to the simplified equation: $$ \mathrm{Hb}(a q)+\mathrm{O}_{2}(a q) \stackrel{k}{\longrightarrow} \mathrm{HbO}_{2}(a q) $$ where the second-order rate constant is \(2.1 \times 10^{6} / M \cdot \mathrm{s}\) at \(37^{\circ} \mathrm{C}\). For an average adult, the concentrations of \(\mathrm{Hb}\) and \(\mathrm{O}_{2}\) in the blood at the lungs are \(8.0 \times 10^{-6} \mathrm{M}\) and \(1.5 \times 10^{-6} M,\) respectively. (a) Calculate the rate of formation of \(\mathrm{HbO}_{2}\). (b) Calculate the rate of consumption of \(\mathrm{O}_{2}\). (c) The rate of formation of \(\mathrm{HbO}_{2}\) increases to \(1.4 \times 10^{-4} M / \mathrm{s}\) during exercise to meet the demand of the increased metabolism rate. Assuming the \(\mathrm{Hb}\) concentration to remain the same, what must the oxygen concentration be to sustain this rate of \(\mathrm{HbO}_{2}\) formation?

What are the units of the rate constant for a third-order reaction?

On which of the following properties does the rate constant of a reaction depend: (a) reactant concentrations, (b) nature of reactants, (c) temperature?

Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right),\) commonly called table sugar, undergoes hydrolysis (reaction with water) to produce fructose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) and glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right):\) $$ \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} $$ \(\begin{array}{ll}\text { fructose } & \text { glucose }\end{array}\) This reaction is of considerable importance in the candy industry. First, fructose is sweeter than sucrose. Second, a mixture of fructose and glucose, called invert sugar, does not crystallize, so the candy containing this sugar would be chewy rather than brittle as candy containing sucrose crystals would be. (a) From the following data determine the order of the reaction. (b) How long does it take to hydrolyze 95 percent of sucrose? (c) Explain why the rate law does not include \(\left[\mathrm{H}_{2} \mathrm{O}\right]\) even though water is a reactant. $$ \begin{array}{cc} \text { Time (min) } & {\left[\mathbf{C}_{12} \mathbf{H}_{22} \mathbf{O}_{11}\right](\boldsymbol{M})} \\ \hline 0 & 0.500 \\ 60.0 & 0.400 \\ 96.4 & 0.350 \\ 157.5 & 0.280 \end{array} $$

The rate law for the reaction: $$ 2 \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ is rate \(=k\left[\mathrm{H}_{2}\right][\mathrm{NO}]^{2}\). Which of the following mechanism can be ruled out on the basis of the observed rate expression? Mechanism I $$ \begin{array}{cc} \mathrm{H}_{2}+\mathrm{NO} \longrightarrow & \mathrm{H}_{2} \mathrm{O}+\mathrm{N} & (\text { slow }) \\ \mathrm{N}+\mathrm{NO} \longrightarrow & \mathrm{N}_{2}+\mathrm{O} & \text { (fast) } \\ \mathrm{O}+\mathrm{H}_{2} \longrightarrow & \mathrm{H}_{2} \mathrm{O} & \text { (fast) } \end{array} $$ Mechanism II $$ \begin{array}{ll} \mathrm{H}_{2}+2 \mathrm{NO} \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} & \text { (slow) } \\ \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \longrightarrow \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O} & \text { (fast) } \end{array} $$ Mechanism III $$ 2 \mathrm{NO} \rightleftarrows \mathrm{N}_{2} \mathrm{O}_{2} $$ (fast equilibrium) $$ \begin{array}{c} \mathrm{N}_{2} \mathrm{O}_{2}+\mathrm{H}_{2} \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} & \text { (slow) } \\ \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \longrightarrow \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O} & \text { (fast) } \end{array} $$
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