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vladimir2022 [97]
3 years ago
14

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.

Engineering
1 answer:
Archy [21]3 years ago
4 0

Answer:

the time, in hours = 4.07hrs

Explanation:

The detailed step by step derivation and appropriate integration is as shown in the attached files.

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The diffusion coefficients for species A in metal B are given at two temperatures:
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a) 149 kJ/mol, b) 6.11*10^-11 m^2/s ,c) 2.76*10^-16 m^2/s

Explanation:

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D = D_0e^{\frac{-Q_d}{RT} }

I will be using R in the equation instead of k_b as the problem asks for molar activation energy

I will be using

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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:

ln(6.69*10^{-17})=ln(D_0) -\frac{Q_d}{R*(1030+273)}

and

ln(6.56*10^{-16}) = ln(D_0) -\frac{Q_d}{R*(1290+273)}

You might notice that these equations have the form of  

d=y-ax

You can solve this equation system easily using calculator, and you will eventually get

D_0 =6.11*10^{-11}\ m^2/s\\ Q_d=1.49 *10^3\ J/mol

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

6.11*10^{-11}e^{\frac{149\ 000}{8.143*(1180+273)}

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