Question

In: Chemistry

Part A The activation energy of a certain reaction is 39.4kJ/mol . At 30 ?C ,...

Part A

The activation energy of a certain reaction is 39.4kJ/mol . At 30 ?C , the rate constant is 0.0190s?1. 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 0.0190s?1 at an initial temperature of 30 ?C , what would the rate constant be at a temperature of 150 ?C for the same reaction described in Part A?

Express your answer with the appropriate units.

Expert Solution

Part A

We're gonna need the Arrhenius equation shown as equation 1 below. I'm gonna try to minimize using equations to avoid confusion...

k1 refers to the rate constant given as 0.0190; I used a subscript 1 because this value changes with temperature T1 (given as 30 deg C but this one should be converted to Kelvin - 303 K).

A is called the frequency factor and varies slightly with small changes in temperature hence we can assume this as constant. Ea refers to the activation energy which is constant as well and is given as 39.4 kJ/mol but we need to convert this to.... 39400 J/mol when using the equation 1. R is the gas constant 8.31 J/K-mol.

Rearranging equation 1 to compute for A (frequency factor):

Now that we have A, we can now determine the required - temperature in deg C when reaction rate is twice.

Using equation 2 below.

taking natural logarithm to both sides of equation 2 would give us this equation 3:

k2 is just simply twice of k1 as mentioned in the problem. k2 = 2*k1 = 2*0.0190 = 0.038

A we assumed constant = 118712.26

Ea we assumed constant = 39400

R = 8.31

Rearranging and solving equation 3:

The answer to Part A is:

Part B

We can use the general equation:

with constant values of A and Ea

T is now 150 deg C = 423 K

Using the equation..

Final answer is:

queen_honey_blossom answered 2 years ago

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