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Chapter 12: Problem 76

Chemists commonly use a rule of thumb that an increase of 10 \(\mathrm{K}\) in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 25 to \(35^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The activation energy for this reaction, considering that the rate of reaction doubles every 10 K increase in temperature, is approximately \(52.6\thinspace kJ\thinspace mol^{-1}\).

Step by step solution

01

Write down the Arrhenius equation

The Arrhenius equation is given by: \(k = A \cdot e^{-\frac{E_a}{RT}}\) where \(k\) is the rate constant, \(A\) is the pre-exponential factor, \(E_a\) is the activation energy, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin.
02

Set up equations for doubling the rate constant

We know that the rate constant doubles every 10 K increase in temperature, so we can set up two equations with \(T_1 = 298K\) and \(T_2 = 308K\) (temperature conversions from Celsius to Kelvin). Our equations will be: \(k_1 = A \cdot e^{-\frac{E_a}{R \cdot T_1}}\) \(k_2 = 2 \cdot k_1 = A \cdot e^{-\frac{E_a}{R \cdot T_2}}\) This step is crucial to show the relationship between temperatures and rate constants according to the problem statement.
03

Divide the second equation by the first equation

Dividing the second equation by the first equation will allow us to eliminate the pre-exponential factor \(A\) and solve for the activation energy \(E_a\). The equations will look like this: \(\frac{k_2}{k_1} = \frac{A \cdot e^{-\frac{E_a}{R \cdot T_2}}}{A \cdot e^{-\frac{E_a}{R \cdot T_1}}}\)
04

Simplify the equation and isolate E_a

Now we can simplify the equation as follows: \(2 = \frac{e^{-\frac{E_a}{R \cdot T_2}}}{e^{-\frac{E_a}{R \cdot T_1}}}\) Taking natural logarithm on both sides, we get: \(ln\thinspace 2 = ln\thinspace(e^{\frac{E_a}{R \cdot T_1} -\frac{E_a}{R \cdot T_2}}) \Longrightarrow ln\thinspace 2 = \frac{E_a}{R \cdot T_1} -\frac{E_a}{R \cdot T_2}\) Now isolating E_a we get: \(E_a = R \cdot \frac{T_1\thinspace T_2}{T_2 - T_1} \cdot ln\thinspace 2\)
05

Substitute the values and solve for E_a

Finally, we can substitute the values of \(T_1\), \(T_2\), and \(R\) into the formula to calculate the activation energy: \(E_a = (8.314\thinspace J\thinspace mol^{-1}K^{-1}) \cdot \frac{(298K)(308K)}{308K - 298K} \cdot ln\thinspace 2\) By calculating the above expression, we obtain: \(E_a \approx 52.6\thinspace kJ\thinspace mol^{-1}\) Thus, the activation energy for this reaction is approximately \(52.6\thinspace kJ\thinspace mol^{-1}\).

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Most popular questions from this chapter

In the Haber process for the production of ammonia, $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ what is the relationship between the rate of production of ammonia and the rate of consumption of hydrogen?

Experiments during a recent summer on a number of fireflies (small beetles, Lampyridaes photinus) showed that the average interval between flashes of individual insects was 16.3 \(\mathrm{s}\) at \(21.0^{\circ} \mathrm{C}\) and 13.0 \(\mathrm{s}\) at \(27.8^{\circ} \mathrm{C}\) a. What is the apparent activation energy of the reaction that controls the flashing? b. What would be the average interval between flashes of an individual firefly at \(30.0^{\circ} \mathrm{C} ?\) c. Compare the observed intervals and the one you calculated in part b to the rule of thumb that the Celsius temperature is 54 minus twice the interval between flashes.

Each of the statements given below is false. Explain why. a. The activation energy of a reaction depends on the overall energy change \((\Delta E)\) for the reaction. b. The rate law for a reaction can be deduced from examination of the overall balanced equation for the reaction. c. Most reactions occur by one-step mechanisms.

A certain reaction has the following general form: $$ \mathrm{aA} \longrightarrow \mathrm{bB} $$ At a particular temperature and \([\mathrm{A}]_{0}=2.80 \times 10^{-3} M,\) con- centration versus time data were collected for this reaction, and a plot of 1\(/[\mathrm{A}]\) versus time resulted in a straight line with a slope value of \(+3.60 \times 10^{-2} \mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\) . a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction. b. Calculate the half-life for this reaction. c. How much time is required for the concentration of A to decrease to $7.00 \times 10^{-4} M ?$

The activation energy for the reaction $$ \mathrm{NO}_{2}(g)+\mathrm{CO}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g) $$ is 125 \(\mathrm{kJ} / \mathrm{mol}\) , and \(\Delta E\) for the reaction is $-216 \mathrm{kJ} / \mathrm{mol}$ . What is the activation energy for the reverse reaction $\left[\mathrm{NO}(g)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{CO}(g)\right] ?$
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