Chemistry kenitics problem

  • Thread starter marcuss
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1. A certain reaction has an activation energy of 54.0 kJ/mol. As the temperature is increased from 22degrees celsius to a higher temperature, the rate constant increases by a factor of 7.00. Calculate the higher temperature.

2. ln(k)=-Ea/RT + ln(A)



3. i dont think that i have enough information because when i plugged the values given and there are still two variables. Then i tried simultanious equations by solving for [A] and there are still two variables the new temp and the rate constant. So if anyone can help it would be very appreciated.
 

Answers and Replies

  • #2
The rate constant is just 7 times the original rate constant, so can't you relate the two together as K1 and 7.K1, and then if you solve for T2 you should be able to get the K1s and the As to cancel out.
 

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