A liquid phase reaction, A → B, is to be carried out in an isothermal, well mixed batch reactor with a volume of 1L. Initially t
here are 6 moles of A. The rate of destruction of A is given by –rA =k1CA/ (1+k2CA), where k1=4, k2 =5. The unit of time in the rate constants is hours. Calculate the time, in hours, that the reaction must proceed in the reactor in order to result in 3 moles of A remaining in the reactor.
1 answer:
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Answer:
a) 149 kJ/mol, b) 6.11*10^-11 m^2/s ,c) 2.76*10^-16 m^2/s
Explanation:
Diffusion is governed by Arrhenius equation
I will be using R in the equation instead of k_b as the problem asks for molar activation energy
I will be using
and
°C + 273 = K
here, adjust your precision as neccessary
Since we got 2 difusion coefficients at 2 temperatures alredy, we can simply turn these into 2 linear equations to solve for a) and b) simply by taking logarithm
So:
and
You might notice that these equations have the form of
You can solve this equation system easily using calculator, and you will eventually get
After you got those 2 parameters, the rest is easy, you can just plug them all including the given temperature of 1180°C into the Arrhenius equation
And you should get D = 2.76*10^-16 m^/s as an answer for c)