250+ TOP MCQs on Rate Data Collection & Analysis – Method of Half Lives and Answers

Chemical Reaction Engineering Questions and Answers for Aptitude test focuses on “Rate Data Collection & Analysis – Method of Half Lives”.

1. State true or false.
For a second order reaction, t_.0.5 is directly proportional to the initial concentration.
A. True
B. False
View Answer

Answer: B
Explanation: The general expression relating t_.0.5, rate constant and initial concentration is t_.0.5 = (frac{C_{A0}^{1-n}(2^{n-1}-1)}{(n-1)k}.) For a second order reaction, t_.0.5 = (frac{1}{kC_{A0}}.) Hence, t_.0.5 is inversely proportional to the initial concentration.

2. In terms of conversion, half life is the time taken for ____
A. 50% conversion
B. 75% conversion
C. 20% conversion
D. 30% conversion
View Answer

Answer: A
Explanation: Half Life is the time to reduce the reactant concentration to half of initial value. Hence, 50% of the reactant is converted to product.

3. For all orders of reaction except the first order, half life is related to initial concentration as ____
A. t_0.5 α C_Aoⁿ
B. t_0.5 α C_Ao^1-n
C. t_0.5 α C_Ao²ⁿ
D. t_0.5 α C_Ao
View Answer

Answer: B
Explanation: The generalised equation is t_0.5 = (frac{C_{A0}^{1-n}(2^{n-1}- 1)}{(n-1)k}.) At half life time, for a reaction of order n, C_A = 0.5C_Ao.

5. The relationship between t_.0.5 and C_Ao for a zero order reaction is ____
A. t_0.5 = (frac{1}{kC_{A0}} )
B. t_0.5 = (frac{1}{C_{A0}} )
C. t_0.5 = (frac{C_{A0}}{k})
D. t_0.5 = (frac{C_{A0}}{2k})
View Answer

Answer: D
Explanation: For zero order reaction, (frac{-dC_A}{dt}) = k. t_0.5 = (frac{C_{A0}- frac{C_{A0}}{2}}{k}.)
Hence, t_0.5 = (frac{C_{A0}}{2k}.)

6. For a zero order reaction, the slope of t_0.5 and C_Ao is ____
A. 2
B. 3
C. 1
D. 0
View Answer

Answer: C
Explanation: Slope is (1-n). At n=0, slope = 1.

7. What is the slope of the plot of ln (t_0.5) and ln (C_Ao)? (Where n is the reaction order)
A. 1-n
B. 2-n
C. n
D. (frac{n}{2} )
View Answer

Answer: A
Explanation: The plot is a straight line of slope (1-n). The equation representing the relationship is ln (t_0.5) = ln((frac{(2^{n-1}-1)}{(n-1)k})) + (1 – n) ln (C_Ao).

8. If t_0.5 = 100 minutes and initial concentration is 10, then the order of the reaction is ____
A. 4
B. -1
C. -2
D. 1
View Answer

Answer: B
Explanation: t_0.5 = (C_{A0}^{1-n}.)100 = 10^{(1 – n)}. Hence, n = -1.

9. The fractional conversion is expressed as____
A. (frac{C_A}{C_{A0}} )
B. (frac{C_{A0}}{C_A} )
C. (frac{1}{C_{A0}} )
D. (frac{C_A}{kC_{A0}} )
View Answer

Answer: A
Explanation: Fractional conversion is the time required to reduce the concentration to any fraction. It is the ratio of final to initial concentration.

10. The time required to convert to a fraction of 0.7 for a first order reaction is ____
A. t_0.7 = (frac{(0.7)}{C_{A0} k})
B. t_0.7 = (frac{(-0.7)}{C_{A0} k})
C. t_0.7 = (frac{(0.3)}{C_{A0} k})
D. t_0.7 = (frac{(-0.3)}{C_{A0} k})
View Answer

Answer: D
Explanation: The time for converting to any fraction, F is, t_F = (frac{C_{A0}^{1-n}(F^{n-1}-1)}{(n-1)k}.) At F = 0.7, t_0.7 = (frac{C_{A0}^{1-2}(0.7^1- 1)}{k}.)