The rate of a certain reaction doubles for every \(10^{\circ} \mathrm{C}\) rise in temperature. (a) How much faster does the reaction proceed at \(45^{\circ} \mathrm{C}\) than at \(25^{\circ} \mathrm{C}\) ? (b) How much faster does the reaction proceed at \(95^{\circ} \mathrm{C}\) than at \(25^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
For part (a), the reaction proceeds 4 times faster at 45°C than at 25°C. For part (b), it proceeds 128 times faster at 95°C than at 25°C.

Step by step solution

01

Identify the Temperature Difference for Part (a)

Calculate the difference in temperature between the two given temperatures for part (a). The difference is found by subtracting the initial temperature from the final temperature: 45°C - 25°C = 20°C.
02

Determine the Number of 10°C Intervals for Part (a)

Since the rate doubles for each 10°C rise, divide the temperature difference by 10°C to find the number of intervals: 20°C / 10°C = 2 intervals.
03

Calculate the Rate Increase for Part (a)

For each interval, the rate doubles. This means for 2 intervals, the rate increases by a factor of 2^2 (since it doubles twice). Thus, at 45°C, the reaction is 2^2 or 4 times faster than at 25°C.
04

Identify the Temperature Difference for Part (b)

Calculate the difference in temperature between the two given temperatures for part (b). The difference is found by subtracting the initial temperature from the final temperature: 95°C - 25°C = 70°C.
05

Determine the Number of 10°C Intervals for Part (b)

Since the rate doubles for each 10°C rise, divide the temperature difference by 10°C to find the number of intervals: 70°C / 10°C = 7 intervals.
06

Calculate the Rate Increase for Part (b)

For each interval, the rate doubles. This means for 7 intervals, the rate increases by a factor of 2^7. Thus, at 95°C, the reaction is 2^7 or 128 times faster than at 25°C.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Arrhenius Equation
The Arrhenius equation is a formula that provides deep insight into the effects of temperature on the rates of chemical reactions. It is expressed as:
\[ k = A e^{-\frac{E_a}{RT}} \]
where:\
    \
  • \(k\) is the rate constant of the reaction,\
  • \(A\) is the frequency factor, representing the number of times particles collide with the correct orientation per unit time,\
  • \(E_a\) is the activation energy of the reaction,\
  • \(R\) is the universal gas constant, and\
  • \(T\) is the absolute temperature (in Kelvin).\
\
The equation demonstrates that an increase in temperature leads to an exponential increase in the reaction rate, as seen in our exercise example where the rate doubles for every 10°C rise. This exponential relationship is because higher temperatures increase the number of particles that have enough energy to overcome the activation energy barrier. When explaining how a reaction speeds up as the temperature rises, we often simplify this concept to the rule of thumb presented in the textbook exercise, but the Arrhenius equation gives us a quantitative way to calculate the effect of temperature changes on reaction rates.
Chemical Kinetics
Chemical kinetics is the branch of chemistry that is concerned with the rates of chemical reactions and the mechanisms by which they occur.
Understanding kinetics allows chemists to determine how different conditions, such as temperature, concentration, and the presence of a catalyst affect the speed of a reaction. It relies on measurements of reaction rates, which tell us how quickly reactants are transformed into products.
In our textbook exercise, the rate of a reaction doubling with each 10°C temperature rise is an application of kinetics. It's important to note that not all reactions will have such a simply defined rate increase with temperature, but kinetics gives us the tools to predict and explain these rate changes. The rate at which reactions occur can be quite complex, involving multiple steps, reactant collisions, and activation energies, which are all described more deeply by the principles of kinetics.
Activation Energy
Activation energy, or \(E_a\), is a term in chemistry that represents the minimum amount of energy needed for reactants to transform into products during a chemical reaction.
It's like the initial ‘push’ needed to start a chemical process. In essence, for reactants to successfully react and form products, their molecules must collide with sufficient energy to overcome the activation energy barrier.
This concept is crucial in the context of our exercise because it helps explain why increasing the temperature results in a faster reaction rate. Increasing temperature boosts the energy of molecules involved in the reaction, increasing the likelihood that collisions will have enough energy to overcome the activation energy barrier. However, it’s not just about reaching this threshold - the distribution of molecular energies at any given temperature can also affect how many reactants have sufficient energy, which ties back to the Arrhenius equation and the temperature dependence of the reaction rate.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Hydrogen reacts with nitrogen monoxide to form dinitrogen monoxide (laughing gas) according to the equation: \(\mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) Determine the rate law, the rate constant, and the orders with respect to each reactant from the following data: $$\begin{array}{|c|c|c|c|} \hline \text { \([\mathrm{NO}](M)\) }& 0.30 & 0.60 & 0.60 \\ \hline \text { \(\left[\mathrm{H}_{2}\right](M)\) } & 0.35 & 0.35 & 0.70 \\\ \hline \text { Rate \(\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{s}^{-1}\right)\) } & 2.835 \times 10^{-3} & 1.134 \times 10^{-2} & 2.268 \times 10^{-2} \\ \hline \end{array}$$

=Suppose that the half-life of steroids taken by an athlete is 42 days. Assuming that the steroids biodegrade by a first-order process, how long would it take for \(\frac{1}{64}\) of the initial dose to remain in the athlete's body?

What is the half-life for the decomposition of \(\mathrm{O}_{3}\) when the concentration of \(\mathrm{O}_{3}\) is \(2.35 \times 10^{-6} \mathrm{M}\) ? The rate constant for this second-order reaction is \(50.4 \mathrm{L} \mathrm{mol}^{-1} \mathrm{h}^{-1}\).

The reaction of compound \(A\) to give compounds \(C\) and \(D\) was found to be second-order in \(A\). The rate constant for the reaction was determined to be \(2.42 \mathrm{L} \mathrm{mol}^{-1} \mathrm{s}^{-1} .\) If the initial concentration is \(0.500 \mathrm{mol} / \mathrm{L},\) what is the value of \(\mathrm{t}_{1 / 2} ?\)

Explain why an egg cooks more slowly in boiling water in Denver than in New York City. (Hint: Consider the effect of temperature on reaction rate and the effect of pressure on boiling point.)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free