12th Grade > Chemistry
CHEMICAL KINETICS MCQs
Total Questions : 30
| Page 2 of 3 pages
Answer: Option A. -> 2.303×2×313×27340log3
:
A
K40∘C=Ae−Ea/R×313K0∘C=Ae−Ea/R×2733=e−Ea/R×(1/313−1/273)
On taking logs on both sides, you get the required answer
:
A
K40∘C=Ae−Ea/R×313K0∘C=Ae−Ea/R×2733=e−Ea/R×(1/313−1/273)
On taking logs on both sides, you get the required answer
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.
:
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.
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
:
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
Answer: Option B. -> K2≈0.25K1
:
B
Let the rate constant at 300 K be K1. Temperature decrease of 10∘C, rate constant value becomes approximately half the original value.
Decrease in temperature = 20∘C
The rate constant, K2 at 280 C is K2=0.25K1
K2=0.25K1
:
B
Let the rate constant at 300 K be K1. Temperature decrease of 10∘C, rate constant value becomes approximately half the original value.
Decrease in temperature = 20∘C
The rate constant, K2 at 280 C is K2=0.25K1
K2=0.25K1
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]
:
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]
Answer: Option C. -> 3
:
C
rate =k[A]x[B]y
4=k[2]x2=k[2]yx=2y=1
:
C
rate =k[A]x[B]y
4=k[2]x2=k[2]yx=2y=1
Answer: Option B. -> 1
:
B
t1/2 is constant. Hence order = 1
:
B
t1/2 is constant. Hence order = 1
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.
:
D
Order of a reaction is equal to the sum of thestoichiometric coefficients expressed in a rate law.
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.
:
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.
Answer: Option A. -> 10−2s−1
:
A
K=0.693t12=0.69369.3=10−2s−1
:
A
K=0.693t12=0.69369.3=10−2s−1