The Arrhenius equation shows the relationship between the rate constant k and th

Question

The Arrhenius equation shows the relationship between the rate constant k and the temperature T in kelvins and is typically written as

k=AeEa/RT

where R is the gas constant (8.314 J/molK), A is a constant called the frequency factor, and Eais the activation energy for the reaction.

However, a more practical form of this equation is

lnk2k1=EaR(1T11T2)

which is mathmatically equivalent to

lnk1k2=EaR(1T21T1)

where k1 and k2 are the rate constants for a single reaction at two different absolute temperatures (T1and T2).

Part A

The activation energy of a certain reaction is 46.1 kJ/mol . At 24  C , the rate constant is 0.0150s1. At what temperature in degrees Celsius would this reaction go twice as fast?

Express your answer with the appropriate units.

Part B

Given that the initial rate constant is 0.0150s1 at an initial temperature of 24  C , what would the rate constant be at a temperature of 110.  C for the same reaction described in Part A?

Express your answer with the appropriate units.

Explanation / Answer

Apply:

T1 = 24°C --> 297K @ K1 = K1

T2 = T2°C --> k2 = 2K1

so

ln(K2/K1) = E/R*(1/T1-1/T2)

ln(2/1) = 46100/8.314*(1/(297 - 1/T2)

T2 = -(ln(2) * 8.314/46100 - 1/297 )^-1

T2 = 308.451 K}

T2 = 308.451-273 = 35.451 °C will be twice!

B)

ln(k2/0.015) = 46100/8.314*(1/(24+273) -1/(110+273))

ln(k2) = 0.015* 46100/8.314*(1/(24+273) -1/(110+273))

k2 = exp(0.06288) = 1.06489 1/s

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