Chemical Reaction Engineering Assessment Questions and Answers focuses on “Stoichiometry – Constant – Volume Batch Reaction Systems”.
1. State true or false.
Constant volume system means that the volume of reactor is constant.
A. True
B. False
Answer: B
Explanation: The volume of the reaction mixture remains constant for a constant volume system. The system is incompressible.
2. Which of the following represents constant volume system?
A. C + O2 → CO2
B. CO + H2O → CO2 + H2
C. N2 + 3H2 → 2NH3
D. SO2 + 0.5O2 → SO3
Answer: B
Explanation: Gas phase reactions in which the number of moles of the product is equal to the number of moles of reactant is a constant volume system. In CO + H2O → CO2 + H2, 2 moles of reactants form 2 moles of product.
3. What is the partial pressure of the product related to the total pressure for a constant volume system? (Where pR is the partial pressure of the product, pR0 is the initial partial pressure of the product, P is the total pressure, P0 is the initial pressure of reaction mixture, r is the stoichiometry of the product and ∆n is the difference between the number of moles of the product and the number of moles of the reactant)
A. pR = (frac{r}{∆n}) (P – P0)
B. pR = pR0 – (frac{r}{∆n}) (P – P0)
C. pR = pR0 + (P – P0)
D. pR = pR0 + (frac{r}{∆n}) (P – P0)
Answer: D
Explanation: For the reaction, aA + bB → rR + sS, pR = CRRT. CR is the product concentration. CRRT = pR0 + (frac{r}{∆n}) (P – P0).
4. The desired relationship between the partial pressure of reactant and total pressure for the reaction N2O4 → 2NO2 is _____
A. PA = PA0 – 2(P – P0)
B. PA = 2PA0 – (P – P0)
C. PA = PA0 – (P – P0)
D. PA = PA0 + (P – P0)
Answer: C
Explanation: ∆n = 2-1 =1. pA = pA0 – (frac{a}{∆n}) (P – P0), a is the stoichiometry of the reactant. Hence, PA = PA0 – (P – P0).
5. If the total pressure for reaction is initially is 2 atm and the pressure is changed to 3 atm for the reaction to occur, then the value of (PA – PA0) for the reaction N2O → N2 + 0.5O2 is ____
A. -2
B. 2
C. -1
D. 1
Answer: A
Explanation: ∆n = 0.5
pA = pA0 – (frac{a}{∆n}) (P – P0)
(PA = PA0) = (frac{-1}{0.5})(3-2)
(PA = PA0) = -2.
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