(a)
Interpretation:
The fractional conversion of propane for the given reaction is to be calculated at
Concept introduction:
For a single reaction system, the final moles of each of the components present, can be estimated by the equation:
Here,
Mole fraction
Here,
Equilibrium constant of this reaction from equation 14.28 can be written as:
Where,
Gibb’s free energy in terms of equilibrium constant is written as:
Also, Gibbs free energy is calculated using heat of reaction from the equation given as
Here,
Where,
(a)
Answer to Problem 14.16P
The fractional conversion of propane for the given reaction at
Explanation of Solution
Given information:
The cracking reaction of propane is:
The equilibrium pressure given for this reaction is
The equilibrium conversion at
The equilibrium temperature for this reaction is taken as
From Table C.4 the standard heat of reaction and Gibb’s free energy for this reaction is:
From Table C.1 the coefficients for the heat capacity of the component gases are given as:
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Now, use equations set (6) to evaluate the values of
Now, use equation (5) along with the above calculated values to get the value of
Use this calculated value of
For the given cracking reaction, let the extent of the reaction be
The individual stoichiometric coefficients for all the components in this reaction is:
The overall stoichiometric coefficient for this reaction is,
Using equation (1), write the expressions for final moles of all the components present in the products as gases.
Total moles of the products will be:
Using equation (2) to write the mole fraction of all the species as:
Now, use equation (3) and the calculated value of the equilibrium constant and calculate
Therefore, the fractional conversion of propane is
(b)
Interpretation:
The temperature for the given fractional conversion of propane at
Concept introduction:
For a single reaction system, the final moles of each of the components present, can be estimated by the equation:
Here,
Mole fraction
Here,
Equilibrium constant of this reaction from equation 14.28 can be written as:
Where,
Gibb’s free energy in terms of equilibrium constant is written as:
Also, Gibbs free energy is calculated using heat of reaction from the equation given as
Here,
Where,
(b)
Answer to Problem 14.16P
The temperature for
Explanation of Solution
Given information:
The cracking reaction of propane is:
The equilibrium pressure given for this reaction is
The equilibrium conversion at
The fractional conversion at equilibrium is
The initial and final pressure of the system is kept at
Fractional conversion of propane is given as
The final mole fraction of all the species will be:
Now, use equation (3) to calculate the equilibrium constant of this reaction as:
From Table C.4 the standard heat of reaction and Gibb’s free energy for this reaction is:
From part (a), the calculated values of
Make an initial guess for the equilibrium temperature at
Now, use equations set (6) to evaluate the values of
Now, use equation (5) along with the above calculated values to get the value of
Use this calculated value of
Since, the calculated value of
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Chapter 14 Solutions
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