Sail E0 Webinar

12th Grade > Chemistry

CHEMICAL KINETICS MCQs

Total Questions : 30 | Page 2 of 3 pages
Question 11. Milk turns sour at 40C three times as faster as 0C. Hence, Ea in the process of turning of milk sour is
  1.    2.303×2×313×27340log3
  2.    2.303×2×313×27340log(1/3) cal
  3.    2.303×2×40273×313log3 cal
  4.    None of these
 Discuss Question
Answer: Option A. -> 2.303×2×313×27340log3
:
A
K40C=AeEa/R×313K0C=AeEa/R×2733=eEa/R×(1/3131/273)
On taking logs on both sides, you get the required answer
Question 12. Which one of the following is a second order reaction?
  1.    H2+Br2→2HBr
  2.    NH4NO3→N2+3H2O
  3.    H2+Cl2→2HCl
  4.    CH3COOCH3+NaOH→CH3COONa+CH3OH
 Discuss Question
Answer: Option D. -> CH3COOCH3+NaOH→CH3COONa+CH3OH
:
D
Choices (a) and (c) are photochemical reaction of zero order. Choice (b) is first order reaction. In choice (d), rate of reaction depends on 1st power of CH3COOCH3 and NaOH both.
Question 13. For reaction: A + B Products, the rate if the reaction at various concentration are given below:
Expt. No.(A)(B)Rate10.20.2220.20.4430.60.436
  1.    Rate =k[A]2[B]2
  2.    Rate =k[A]2[B]
  3.    Rate =k[A][B]2
  4.    Rate =k[A]3[B]
 Discuss Question
Answer: Option B. -> Rate =k[A]2[B]
:
B
Let the rate law be r=k[A]x[B]y
Divide (2) by (1)
2=[2]y,y=1
Divide (3) by (2)
9=(3)x,x=2
Hence rate equation, R=k[A]2[B]1
Question 14. A chemical reaction was carried out at 300 K and 280 K. The rate constants were found to be K1 and K2 respectively then  
  1.    K2≈0.5K1
  2.    K2≈0.25K1
  3.    K2≈2K1
  4.    K2≈4K1 
 Discuss Question
Answer: Option B. -> K2≈0.25K1
:
B
Let the rate constant at 300 K be K1. Temperature decrease of 10C, rate constant value becomes approximately half the original value.
Decrease in temperature = 20C
The rate constant, K2 at 280 C is K2=0.25K1
K2=0.25K1
Question 15. The data for the reaction is A + B  C is
 Exp.[A]0[B]0Initial rate10.0120.0350.1020.0240.0350.8030.0120.0700.1040.0240.0700.80
  1.    r=k[B]3
  2.    r=k[A]3
  3.    r=k[A][B]4
  4.    r=k[A]2[B]2
 Discuss Question
Answer: Option B. -> r=k[A]3
:
B
A quick way to solve this would be to look at rows one and two. Straight away you can tell that rate does not depend on [B]
Question 16. For the reaction A + B C, it is found that doubling the concentration of A increases the rate by 4 times and doubling the concentration of B doubles the reaction rate. What is the overall order of the reaction?
  1.    4
  2.    32
  3.    3
  4.    1
 Discuss Question
Answer: Option C. -> 3
:
C
rate =k[A]x[B]y
4=k[2]x2=k[2]yx=2y=1
Question 17. If in the fermentation of sugar in an enzymatic solution which is initially 0.12 M the  concentration of sugar is reduced to 0.06 M in 10 h and to 0.03 M in 20 h, hence order of the reaction is
  1.    0
  2.    1
  3.    2
  4.    3
 Discuss Question
Answer: Option B. -> 1
:
B
t1/2 is constant. Hence order = 1
Question 18. Which of these does not influence the rate of reaction?
  1.    Nature of the reactants
  2.    Concentration of the reactants
  3.    Temperature 
  4.    Order of a reaction
 Discuss Question
Answer: Option D. -> Order of a reaction
:
D
Order of a reaction is equal to the sum of thestoichiometric coefficients expressed in a rate law.
Question 19. One litre of 2M acetic acid is mixed with one litre of 3M ethyl alcohol to form ester CH3COOH+C2H5OHCH3COOC2H5+H2O. The decrease in the initial rate if each solution is diluted by an equal volume of water would be
  1.    0.5 times
  2.    2 times
  3.    4 times
  4.    0.25 times
 Discuss Question
Answer: Option C. -> 4 times
:
C
When each solution is diluted by equal volume of water, the concentration of each chemical is reduced to half the original.
We assume that this is an elementary reaction which leads to the conclusion that the order of this reaction would be 2.
The rate equation is v=k[CH2COOH]1×[C2H5OH]1. Hence the rate decreases by a factor of 12×12=14.
Thus the rate after dilution will be 4 times lesser than earlier rate.
Question 20. The half life period for a first order reaction is 69.3s. Its rate constant is
  1.    10−2s−1
  2.    10−4s−1
  3.    10s−1
  4.    102s−1
 Discuss Question
Answer: Option A. -> 10−2s−1
:
A
K=0.693t12=0.69369.3=102s1

Latest Videos

Latest Test Papers