Home
Class 12
CHEMISTRY
Rates of reaction double with every 10^@...

Rates of reaction double with every `10^@` rise in temperature. If this generalization holds for a reaction in the temperature ranges 298 K to 308 K, what would be the value of activation energy for their reaction ? R = 8.314 J `K^(-1) "mol"^(-1)`.

Text Solution

AI Generated Solution

To determine the activation energy (Ea) for the reaction given that the rate of reaction doubles with every 10°C rise in temperature, we can use the Arrhenius equation and the information provided. ### Step-by-Step Solution: 1. **Understanding the Problem**: - We know that the rate of reaction doubles (k2 = 2k1) when the temperature increases by 10°C. - The initial temperature (T1) is 298 K and the final temperature (T2) is 308 K. ...
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL KINETICS

    CBSE COMPLEMENTARY MATERIAL|Exercise SHORT ANSWER TYPE-II QUESTIONS|23 Videos

  • BIOMOLECULES

    CBSE COMPLEMENTARY MATERIAL|Exercise SHORT ANSWER-II TYPE QUESTION|13 Videos

  • CHEMISTRY IN EVERYDAY LIFE

    CBSE COMPLEMENTARY MATERIAL|Exercise LONG ANSWER TYPE QUESTIONS|2 Videos

Similar Questions

Explore conceptually related problems

In general, it is observed that the rate of a chemical reaction doubles with every 10^(@) rise in temperature. If the generalisation holds for a reaction in the temperature range 295 K to 305 K, what would be the value of activation energy for the reaction?

The rate of a reaction doubles when the tempeature changes from 300 K to 310 K. Activation energy for the reaction is: (R = 8.314 JK^(-1) mol^(-1), log2 = 0.3010)

The rate of a reaction triples when temperature changes from 50 to 100^(@)C. Calculate the energy of activation for such a reaction (R=8.314" J K"^(-1)" mol"^(-1))

The rate of a reaction increases four times when the temperature changes from 300K to 320 K. Calculate the energy of activation of the reaction. (R=8.314JK^(-1)mol^(-1))

The rates of most reactions double when their temperature is raised from 298 K to 308 K. Calculate their activation energy.

The rate of a reaction doubles when its temperature changes form 300 K to 310 K . Activation energy of such a reaction will be: (R = 8.314 JK^(-1) mol^(-1) and log 2 = 0.301)

The rate constant for a first order reaction becomes six times when the temperature is raised from 350 K to 400 K. Calculate the activation energy for the reaction [R=8.314JK^(-1)mol^(-1)]

The rates of most reaction double when their temperature is raised from 298K to 308K . Calculate their activation energy.