If the activation energy of a reaction is \(4.86 \mathrm{~kJ}\), then what is the percent increase in the rate constant of a reaction when the temperature is increased from \(45^{\circ}\) to \(75^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
Answer: The percent increase in the rate constant of the reaction when the temperature is increased from 45°C to 75°C is 71.8%.

Step by step solution

01

Convert temperatures to Kelvin

First, we need to convert the given temperatures from Celsius to Kelvin. To do this, add 273.15 to the Celsius temperature: \(45^{\circ} \mathrm{C} = 45 + 273.15 = 318.15 \mathrm{K}\) \(75^{\circ} \mathrm{C} = 75 + 273.15 = 348.15 \mathrm{K}\)
02

Calculate rate constants using the Arrhenius equation

Recall that the Arrhenius equation relates the rate constant (k) of a reaction to its activation energy (Ea) and temperature (T) as follows: \(k = A e^{\frac{-Ea}{RT}}\) where A is the pre-exponential factor, R is the ideal gas constant (8.314 J / (mol*K)), and T is the temperature in Kelvin. We want to find the rate constants at both temperatures, so we will use the equation twice: \(k_{318.15} = A e^{\frac{-4.86 \times 10^3 \mathrm{J/mol}}{8.314 \mathrm{J/(mol*K)} \times 318.15 \mathrm{K}}}\) \(k_{348.15} = A e^{\frac{-4.86 \times 10^3 \mathrm{J/mol}}{8.314 \mathrm{J/(mol*K)} \times 348.15 \mathrm{K}}}\)
03

Calculate the ratio of rate constants

Now that we have expressions for the rate constants at each temperature, we can find the ratio of rate constants: \(\frac{k_{348.15}}{k_{318.15}} = \frac{A e^{\frac{-4.86 \times 10^3 \mathrm{J/mol}}{8.314 \mathrm{J/(mol*K)} \times 348.15 \mathrm{K}}}}{A e^{\frac{-4.86 \times 10^3 \mathrm{J/mol}}{8.314 \mathrm{J/(mol*K)} \times 318.15 \mathrm{K}}}}\) Notice that the A terms cancel out, leaving: \(\frac{k_{348.15}}{k_{318.15}} = \frac{e^{\frac{-4.86 \times 10^3 \mathrm{J/mol}}{8.314 \mathrm{J/(mol*K)} \times 348.15 \mathrm{K}}} }{e^{\frac{-4.86 \times 10^3 \mathrm{J/mol}}{8.314 \mathrm{J/(mol*K)} \times 318.15 \mathrm{K}}}}\)
04

Calculate the percent increase in rate constants

Finally, we can determine the percent increase in the rate constant as follows: Percent Increase \(= \frac{k_{348.15} - k_{318.15}}{k_{318.15}} \times 100 = \frac{k_{348.15}}{k_{318.15}} \times 100 - 100\) Use a calculator to compute the exponential terms and the ratio of rate constants, which yields: Percent Increase \(= \frac{1.718}{1} \times 100 - 100 = 71.8 \%\) Thus, the percent increase in the rate constant of the reaction when the temperature is increased from \(45^{\circ}\) to \(75^{\circ} \mathrm{C}\) is 71.8%.

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Key Concepts

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

Activation Energy
Activation energy, denoted as Ea, is a critical concept in understanding chemical reactions. It represents the minimum amount of energy required for reactants to transform into products during a chemical reaction. In essence, only when the molecules have enough energy to overcome this energy barrier can a reaction proceed.

Think of it as a hill that reactants must climb to convert into products; the higher the hill (activation energy), the fewer the molecules that will have enough energy to reach the top and react. This is why reactions with high activation energies happen slower under the same conditions compared to those with lower activation energies. To visualize activation energy, imagine a match that won't ignite without sufficient friction; that friction provides the energy needed to start the reaction. In the context of the step-by-step solution provided, the activation energy of the reaction is given as 4.86 kJ, and it plays a crucial role in calculating the rate constant change with temperature.
Rate Constant
The rate constant of a reaction, often denoted by the symbol k, is a proportionality constant that links the reaction rate to the reactant concentrations as described by the rate law. Its value is determined by the specific chemical reaction and the conditions under which the reaction takes place, such as temperature.

A pivotal aspect of the rate constant is that it changes with temperature, which is well captured by the Arrhenius equation. This equation illustrates how the rate constant k increases with an increase in temperature, provided that other factors remain unchanged. In mathematical terms, the equation quantifies how the rate constant is affected by the activation energy and the temperature. As observed in the solution steps, the rate constant k at different temperatures can be determined using the given activation energy and the Arrhenius equation. After calculating the rate constants at two different temperatures, the percent increase between them demonstrates how sensitive the rate constant is to temperature changes.
Temperature and Reaction Rate
Temperature plays a significant role in the rate of a chemical reaction. An increase in temperature generally causes an increase in the reaction rate. This happens because higher temperatures mean that molecules move faster, and as a result, collide more frequently and with greater energy.

These more energetic collisions increase the likelihood that the molecules will have enough energy to overcome the activation energy barrier, leading to more successful reactions per unit time. The relationship between temperature and reaction rate is not linear but is well described by the Arrhenius equation, showing an exponential dependence. As illustrated in the exercise, raising the temperature from 318.15 K to 348.15 K resulted in a significant increase in the rate constant, and thus in the reaction rate itself, indicating that even small temperature changes can have a large impact on the rate of reaction. This principle is crucial in many practical applications, such as cooking, where adjusting the temperature can alter how quickly food cooks, or in industrial processes where precise control over reaction rates is essential for efficiency and safety.

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Most popular questions from this chapter

The decomposition of phosphine, \(\mathrm{PH}_{3}\), to \(\mathrm{P}_{4}(g)\) and \(\mathrm{H}_{2}(g)\) is firstorder. Its rate constant at a certain temperature is \(1.1 \mathrm{~min}^{-1}\). (a) What is its half-life in seconds? (b) What percentage of phosphine is decomposed after \(1.25 \mathrm{~min}\) ? (c) How long will it take to decompose one fifth of the phosphine?

Copper-64 is one of the metals used to study brain activity. Its decay constant is \(0.0546 \mathrm{~h}^{-1}\). If a solution containing \(5.00 \mathrm{mg}\) of \(\mathrm{Cu}-64\) is used, how many milligrams of Cu-64 remain after eight hours?

Experimental data are listed for the hypothetical reaction $$\mathrm{x} \longrightarrow \mathrm{Y}+\mathrm{Z}$$ $$\begin{array}{lcccccc}\hline \text { Time (s) } & 0 & 10 & 20 & 30 & 40 & 50 \\\ {[\mathrm{X}]} & 0.0038 & 0.0028 & 0.0021 & 0.0016 & 0.0012 & 0.00087 \\\\\hline\end{array}$$ (a) Plot these data as in Figure \(11.2 .\) (b) Draw a tangent to the curve to find the instantaneous rate at \(40 \mathrm{~s}\). (c) Find the average rate over the 10 to 50 s interval. (d) Compare the instantaneous rate at \(40 \mathrm{~s}\) with the average rate over the \(40-\mathrm{s}\) interval.

WEB When boron trifluoride reacts with ammonia, the following \(T\) reaction occurs: for $$\mathrm{BF}_{3}(g)+\mathrm{NH}_{3}(g) \longrightarrow \mathrm{BF}_{3} \mathrm{NH}_{3}(g)$$ The following data are obtained at a particular temperature: $$\begin{array}{cccc}\hline \text { Expt. } & {\left[\mathrm{BF}_{3}\right]} & {\left[\mathrm{NH}_{3}\right]} & \text { Initial Rate }(\mathrm{mol} / \mathrm{L} \cdot \mathrm{s}) \\\\\hline 1 & 0.100 & 0.100 & 0.0341 \\ 2 & 0.200 & 0.233 & 0.159 \\ 3 & 0.200 & 0.0750 & 0.0512 \\ 4 & 0.300 & 0.100 & 0.102 \\\\\hline\end{array}$$

The chirping rate of a cricket \(\mathrm{X}\), in chirps per minute near room temperature, is $$\mathrm{X}=7.2 t-32$$ where \(t\) is the temperature in \({ }^{\circ} \mathrm{C}\). (a) Calculate the chirping rates at \(25^{\circ} \mathrm{C}\) and \(35^{\circ} \mathrm{C}\). (b) Use your answers in (a) to estimate the activation energy for the chirping. (c) What is the percentage increase for a \(10^{\circ} \mathrm{C}\) rise in temperature?

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