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Ch.14 - Chemical Kinetics

Chapter 14, Problem 88

When the temperature of a gas is raised by 10 °C, the collision frequency increases by only about 2%, but the reaction rate increases by 100% (a factor of 2) or more. Explain.

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Hello. In this problem we are told a 20°C rise in gas temperature results in a 3% increase in collision frequency. Yet a factor for more increase in the rate of reaction. And were asked to explain this observation, let's make use of the equation to explain what's been stated in this problem. So the rennes equation recall, we have our reaction rate constant is equal to a pre exponential factor. A. And then we have the exponents of the negative of our activation energy divided by the gas constant divided by temperature. This pre exponential factor then can be written as to the product to other factors. So Z is the frequency factor and P is the steering factor. So they ran this equation can make sense in terms of collision theory. According to collision theory. Then in order for a chemical reaction to occur, reacting molecules have to collide. So this tells us the frequency of collusion. They have to collide with proper orientation. So the steering factor tells us what proportion of those molecules that collide have the proper orientation and they have to collide with sufficient energy which is given by the activation energy. The frequency factors e. Is proportional to temperature. So as temperature increases, We increase the average kinetic energy of the gas molecules moving around more quickly, which means that there are more likely to glide. So we're going to increase than the frequency of collision. And we're told then at a 20°C rise in gas temperature would result in a 3% increase in the collision frequency. Now, considering the other portion of this question, we will make use of the exponential portion of the chinese equation. So this is frequently referred to as just lowercase F. Which then is the fraction of collisions with adequate energy to overcome the energy barrier. And so that means then this is the fraction of collisions that have energy equal to or greater than the activation energy. As we increase the temperature can this will result in a increase in the average connected energy as reacting molecules are moving around more quickly. We will increase the frequency of collision and we will also increase the forcefulness of collision, which means then we will have an increase in the fraction of collisions with adequate energy to overcome our energy barrier. And this is going to increase exponentially. We are dealing with again the exponential portion of the Iranians equation. So then to summarize as the temperature increases, the fraction of collisions with adequate energy to go on to react increases exponentially. And so this corresponds to then answer a thanks for watching. Hope this help
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The values of Ea = 248 kJ>mol and ΔE = 41 kJ>mol have been measured for the reaction H21g2 + CO21g2S H2O1g2 + CO1g2 (b) Considering the geometry of the reactants and products, suggest a plausible structure for the transition state.
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Textbook Question

Consider three reactions with different values of Ea and ΔE:

Reaction 1. Ea = 20 kJ>mol; ΔE = -60 kJ/mol

Reaction 2. Ea = 10 kJ>mol; ΔE = -20 kJ/mol

Reaction 3. Ea = 40 kJ>mol; ΔE = +15 kJ/mol

(b) Assuming that all three reactions are carried out at the same temperature and that all three have the same frequency factor A, which reaction is the fastest and which is the slowest?

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