Activation energy...Help with #3 please. Thanks! Equation to be used is Ea = R (

Question

Activation energy...Help with #3 please. Thanks!

Equation to be used is Ea = R ( (ln k1/k2) / 1/T2 - 1/T1); R is 8.314 and answer should be in J/mol. I only need number 3. What should I do about the "doubles the rate constant"?

2) The chirping of tree crickets has sometimes been used to predict temperature. At 25.0 degree C the rate of chirping was recorded to be 179 chirps per minute. At 21.9 degree. It was recorded to be 142 chirps/min. Calculate the activation energy for this reaction. 3) If a temperature increase from 10.0 degree C to 20.0 degree C doubles the rate constant for a reaction, what is the value of the activation energy for the reaction? 4) Increasing the temperature of a reaction from 20.0 degree C to an unknown temperature triples the rate constant for a reaction. Given that the activation energy for the reaction is 54.95 kJ/mol, determine the unknown temperature in degree C.

Explanation / Answer

2) log(k2/k1) =( Ea/2.303R)[1/T1-1/T2]

     log(179/142) =( Ea/2.303x2)[1/294-1/298)

Ea= 10383.7cal

                      

3) log(k2/k1) =( Ea/2.303R)[1/T1-1/T2]

     log 2 =( Ea/2.303R)[1/283-1/293]

           Ea = 11553cal

4) log(k2/k1) =( Ea/2.303R)[1/T1-1/T2]

    log 3 = (0 .5495/2.303x 8.314)[1/T1-1/293]

          T1 = 273k

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