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\(\frac{k_{35^{\circ}}}{k_{34^{\circ}}}>1\), this means that (a) Rate increases with the rise in temperature (b) Rate decreases with rise in temperature (c) rate does not change with rise in temperature (d) None of the above

Short Answer

Expert verified
The rate of reaction increases with the rise in temperature, so the correct answer is (a) Rate increases with the rise in temperature.

Step by step solution

01

Understanding the Given Inequality

The inequality \(\frac{k_{35^\circ}}{k_{34^\circ}}>1\) represents a comparison between the rate constants (k) at two different temperatures, 35°C and 34°C. If the ratio is greater than 1, it means the rate constant at the higher temperature is larger.
02

Analyzing the Meaning of a Larger Rate Constant

A larger rate constant implies a higher reaction rate. Therefore, if \(k_{35^\circ}>k_{34^\circ}\), the reaction rate at 35°C is faster than at 34°C.
03

Concluding the Temperature's Effect on Rate

Since the rate constant increases with a rise in temperature, it can be inferred that the reaction rate also increases with the rise in temperature.

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Key Concepts

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

Temperature Dependence of Reaction Rates
Understanding how temperature affects reaction rates is crucial for predicting how a chemical reaction will proceed under different conditions. Essentially, a fundamental principle in chemistry is that, as temperature increases, the reaction rate generally does too.

This principle is rooted in the kinetic molecular theory, where higher temperature translates to greater kinetic energy in the reacting molecules. Consequently, molecules move faster, collide more frequently, and do so with more energy, leading to an increased rate of successful collisions—that is, collisions that can overcome the activation energy barrier and lead to product formation. This is precisely what is observed in the given exercise where the rate constant, a direct measure of reaction rate, is higher at 35°C compared to 34°C, thus implying a quicker reaction at the higher temperature.

For a student trying to grasp this concept, it's important to visualize molecules as being more 'energetic' at higher temperatures, leading to a greater likelihood they'll react. Simple analogies, like people being more active on warmer days compared to colder ones, might help in understanding this temperature-reaction rate relationship.
Rate Constant Comparison
Comparing rate constants at different temperatures allows chemists to understand how a slight change in temperature can affect the speed of a reaction. The rate constant, denoted as 'k', is a proportionality constant in the rate equation of a chemical reaction. It ties the reaction rate to the concentration of reactants.

When the rate constant increases, it signifies a boost in reaction rate under given conditions. In the exercise provided, we observe the ratio of rate constants \(\frac{k_{35^\circ}}{k_{34^\circ}} > 1\), indicating that the rate constant and, thus, the reaction rate, is higher at 35°C than at 34°C. This knowledge is practical not just in academic exercises but also in industrial applications where achieving optimal reaction rates can be critical for efficiency and safety.

Students must be mindful of the fact that while the relationship between temperature and reaction rate is semi-qualitative in nature, precise quantitative analysis requires understanding the rate constants' temperature dependency through mathematical models like the Arrhenius equation.
Arrhenius Equation
The Arrhenius equation is a formula that provides a quantitative basis for the relationship between temperature and reaction rate. The equation is expressed as \(k = A e^{-\frac{E_a}{RT}}\), where \(k\) is the rate constant, \(A\) is the pre-exponential factor that represents the frequency of collisions, \(E_a\) is the activation energy, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin.

This powerful equation shows how the rate constant changes exponentially with temperature. A small increase in temperature can significantly decrease the exponential term, thus, increasing the rate constant. Solving this equation at different temperatures allows chemists to predict reaction rates and understand reaction mechanisms.

For a more accessible understanding, students should think of the Arrhenius equation as explaining why even small temperature increases can lead to big jumps in reaction speed - because of exponential changes dictated by activation energy, akin to a 'reactivity accelerator'. Mastery of this concept is crucial for anyone looking to work with chemical reactions, from the classroom to the laboratory.

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Most popular questions from this chapter

The unit of rate constant of zero order and first order chemical reactions are respectively: (a) \(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}, \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) (b) \(\mathrm{s}^{-1}, \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~s}^{-1}\) (c) \(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}, \mathrm{~s}^{-1}\) (d) None of these

The rate constant of a reaction depends on (a) temperature (b) pressure (c) extent of reaction (d) initial concentration of the reactant

For a first order homogeneous gaseous reaction, \(A \longrightarrow 2 B+C\) then initial pressure was \(P_{i}\) while total pressure after time ' \(t\) ' was \(P_{t}\). The right expression for the rate constants \(k\) in terms of \(P_{i}, P_{t}\) and \(t\) is : (a) \(k=\frac{2.303}{t} \log \left(\frac{2 P_{i}}{3 P_{i}-P_{t}}\right)\) (b) \(k=\frac{2.303}{t} \log \left(\frac{2 P_{i}}{2 P_{t}-P_{i}}\right)\) (c) \(k=\frac{2.303}{t} \log \left(\frac{P_{i}}{P_{i}-P_{t}}\right)\) (d) none of these

The rate of reaction is expressed in different ways as follows: \(+\frac{1}{2} \frac{d[C]}{d t}=-\frac{1}{3} \frac{d[D]}{d t}=+\frac{1}{4} \frac{d[A]}{d t}=-\frac{d[B]}{d t}\) The reaction is: (a) \(4 A+B \longrightarrow 2 C+3 D\) (b) \(B+3 D \longrightarrow 4 A+2 C\) (c) \(A+B \longrightarrow C+D^{\prime}\) (d) \(B+D \longrightarrow A+C\)

For the zero order reaction \(A \rightarrow B+C ;\) initial concentration of \(A\) is \(0.1 M\). If \(A=0.08 M\) after 10 minutes, then it's half-life and completion time are respectively: (a) \(10 \mathrm{~min} ; 20 \mathrm{~min}\) (b) \(2 \times 10^{-3} \min ; 4 \times 10^{-3} \min\) (c) \(25 \mathrm{~min}, 50 \mathrm{~min}\) (d) \(250 \mathrm{~min}, 500 \mathrm{~min}\)

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