Activation Energy related to rate.

In summary, the conversation discusses the activation energy for a reaction and how to calculate the temperature at which the rate of the reaction is 80.0 times greater than at 300.0 degrees C. Using the gas constant and Arrhenius equation, the temperature can be determined. The question also clarifies the temperature range in which the reaction was studied.
  • #1
dgoudie
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0
[SOLVED] Activation Energy related to rate.

The activation energy for the reaction : [tex] 2N_{2}O_{5} (g) \rightarrow 2N_{2}O_{4} (g) + O{2} (g),[/tex] is 103 kJ/mol.

The gas constant is 8.314 J/mol*K
Using only information given in this problem calculate the temperature at which the rate of the reaction is 80.0 times greater than it is at 300.0 degrees C (Notice that the given information concerns rates, not rate constants.)


I'm not really sure how do go about this question. do I still use Arrhenius's Equation?

any help would be greatly appreciated. thanks
 
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  • #2
Yes, you can use the Arrhenius equation. Did you really mean 300 C? Are you sure it isn't 300 K? As far as I know, this very complicated first-order reaction was only studied over the temperature range 0 C to 123 C but I may be wrong. Please check.
 
  • #3


I can provide some insights on how to approach this problem. First, let's review the concept of activation energy. Activation energy is the minimum energy required for a chemical reaction to occur. In other words, it is the energy barrier that must be overcome for a reaction to take place.

In this case, the rate of the reaction is directly related to the activation energy. The higher the activation energy, the slower the reaction rate will be. Conversely, a lower activation energy will result in a faster reaction rate.

To calculate the temperature at which the rate of the reaction is 80.0 times greater than at 300.0 degrees C, we can use the Arrhenius equation:

k = A * e^(-Ea/RT)

where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.

Since we are given the activation energy and the gas constant, we can rearrange the equation to solve for the temperature:

T = -Ea / (R * ln(k/A))

Since the rate is 80.0 times greater, we can set up the following equation:

80.0 = k at unknown temperature / k at 300.0 degrees C

Solving for k at unknown temperature and substituting into the rearranged Arrhenius equation, we get:

T = -103 kJ/mol / (8.314 J/mol*K * ln(80.0))

T = 1146.5 K

Therefore, the temperature at which the rate of the reaction is 80.0 times greater than at 300.0 degrees C is approximately 1146.5 Kelvin. I hope this helps!
 

1. What is activation energy?

Activation energy is the amount of energy required to start a chemical reaction. It is the minimum amount of energy that reactant molecules must have in order to undergo a reaction.

2. How does activation energy affect the rate of a reaction?

The higher the activation energy, the slower the rate of reaction. This is because a higher activation energy means that fewer reactant molecules will have enough energy to overcome the energy barrier and form products.

3. How is activation energy related to the temperature of a reaction?

Increasing the temperature of a reaction increases the average kinetic energy of the reactant molecules. This means more molecules will have enough energy to overcome the activation energy barrier, resulting in a higher reaction rate.

4. Can the activation energy of a reaction be lowered?

Yes, the activation energy of a reaction can be lowered by the presence of a catalyst. A catalyst provides an alternate pathway for the reaction to occur, requiring a lower activation energy and therefore increasing the rate of the reaction.

5. How does the concentration of reactants affect activation energy?

A higher concentration of reactants means more collisions between molecules, increasing the chances of successful collisions and therefore lowering the activation energy of the reaction. This results in a higher reaction rate.

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