A liquid phase reaction, A+B à C+D is to be carried out in a well-mixed ideal batch reactor with a constant volume of 10 liters.
The initial concentration of A is 4.71 mol/L, and the initial concentration of B is one half of the initial concentration of A. If the rate of formation of compound D is given by rD = k*CA*CB, with k = 4.5 L/(mol min), calculate the amount of time (in minutes) needed to achieve a conversion of the limiting reactant of 0.86.
1 answer:
Answer:
Explanation:
Given data
- rate of formation of compound D
- rD = k*CA*CB
- Volume V = 10 L
- initial concentration of A = CA0 = 4.71mol/L
- Initial concentration of B = one half of the initial concentration of A
- CB0 = 0.5 x 4.71 mol/L = 2.355 mol/L
- A + B = C + D
- From the stoichiometry of the reaction
- Mol ratio of A : B = 1 : 1
- moles of A > moles of B
- B is limiting reactant
- Rate constant k = 4.5 L/(mol min)
- -rB = rD = k*CA*CB
- -rB = k*CA0(1-CB0*X)*CB0(1-X)
Other detailed steps is as shown in the attached file
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