Exercises
Exercise 1. Sketch a potential energy diagram, properly labeled, for the reaction between nitrogen monoxide and ozone.
NO (g) + O3 (g) → NO2 (g) + O2 (g)
At a given temperature, Ea is 25.0 kJ, and ΔH is -250 kJ. What is Ea for the reverse reaction?
ΔHrxn = Ea(fwd) – Ea(rev)
Ea(rev) = -ΔHrxn + Ea(fwd) = -(-250 kJ) + 25.0 kJ = 275 kJ
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Exercise 2. Plot the data below to determine the activation energy, Ea, and the frequency factor, A.
The slope is equal to -Ea/R where R is 8.314 J/(mol⋅K)
-14792 = -Ea/R and Ea = 1.22 x 104 J = 12.2 kJ
The frequency factor A
ln 27.843. Take the antilog and A = 1.24 x 104 L mol-1 s-1.
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Exercise 3. A first order reaction has a rate constant of 2.3 x 10,-3 s-1 at 35oC. The temperature is increased to 45oC and the reaction rate doubles. What is the activation energy for this reaction?
Convert temperature to K. Then solve the Arrhenius equation for Ea. \(\displaystyle ln\frac{k_2}{k_1}\;=\;-\frac{E_a}{R}\Bigl(\frac{1}{T_2}\;-\;\frac{1}{T_1}\Bigr)\)
\(\displaystyle E_a\;=\;-\frac{ln\frac{k_2}{k_1}\times R}{\Bigl(\frac {1}{T_2}\;-\;\frac{1}{T_1}\Bigr)}\;=\;-\frac{ln\frac{2\times(2.3\times 10^{-3}s^{-1})}{2.3\times 10^{-3}s^{-1}}\times 8.314\frac{J}{mol⋅K}}{\Bigl(\frac {1}{318\;K}\;-\;\frac{1}{308\; K}\Bigr)}\;=\;2132\;J/mol\)
Ea = 2.1 x 103 kJ/mol
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Exercise 4. Consider the following reaction.
2 HI (g) → H2 (g) + I2 (g)
The rate constant at 226.9oC is 9.50 x 10-9 M-1 s-1 and 1.11 x 10-5 M-1 s-1 at 326.9oC. What is the activation energy?
Convert temperature to K. Then solve the Arrhenius equation for Ea. \(\displaystyle ln\frac{k_2}{k_1}\;=\;-\frac{E_a}{R}\Bigl(\frac{1}{T_2}\;-\;\frac{1}{T_1}\Bigr)\)
\(\displaystyle E_a\;=\;-\frac{ln\frac{k_2}{k_1}\times R}{\Bigl(\frac {1}{T_2}\;-\;\frac{1}{T_1}\Bigr)}\;=\;-\frac{ln\frac{1.11\times 10^{-5}M^{-1}s^{-1}}{9.58\times 10^{-9}M^{-1}s^{-1}}\times 8.314\frac{J}{mol⋅K}}{\Bigl(\frac {1}{600.05\;K}\;-\;\frac{1}{500.05\; K}\Bigr)}\;=\;1.76\times 10^5\;J/mol\)
Ea = 176 kJ/mol
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Exercise 5. Using the following energy diagram, determine Ea(forward), Ea(reverse), the transition state, and ΔHrxn. Draw the transition state.
Ea(for) = 78 kJ, Ea(rev) = 6 kJ, and ΔHrxn = 72 kJ
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Exercise 6. Calculate the number of effective collisions out of 10 billion collisions for each of the following:
Calculate f, the fraction of collisions with enough energy for reaction to occur.
This means, if we have 10 billion, 1010, collisions only
\(\displaystyle \frac{x}{10^{10}}\;=\;2.49\times 10^{-13}\)
1010 x (2.49 x 10-13) = 0.00249 collisions out of 10 billion would lead to a reaction.
b) the same reaction with Ea = 24 kJ at a temperature of 98oC.
This means, if we have 10 billion, 1010, collisions only
\(\displaystyle \frac{x}{10^{10}}\;=\;4.18\times 10^{-4}\)
1010 x (4.18 x 10-4) = 4.18 x 106 collisions out of 10 billion would lead to a reaction.
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