Chemical Kinetics Test

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Chemical Kinetics Test - 3

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Weekly Quiz Competition

Question 1
1 / -0

The order of the reaction whose rate and rate constant are the same is

A
1

B
0

C
2

D
4

Solution

For zero order reaction, both are the same. Also, it is true that units for rate and rate constant are the same for zero order.
Question 2
1 / -0

Activation energy of a chemical reaction can be determined by plotting

A
logarithm of rate constant (y-axis) against reciprocal of absolute temperature (x-axis)

B
logarithm of rate constant (y-axis) against time (x-axis)

C
rate constant (y-axis) against reciprocal of time (x- axis)

D
logarithm of rate constant (y-axis) against logarithm of reciprocal of absolute temperature (x-axis)

Solution

Activation energy of a chemical reaction can be found by applying Arrhenius equation. So, this options is correct.
Question 3
1 / -0

2A B + C

For the above reaction - d[A]/dt = k[A]²
The half life of the reactant is

A
1/k

B
1/k[A]³

C
1/k[A]²

D
1/k[A]

Solution

This will be appropriate half life period for second order and the given reaction is of second order.
Question 4
1 / -0

The rate of a reaction is r₁ at 35^oC. The rate of a reaction at 45^oCwill be

A
r₁/2

B
2 r₁

C
10 r₁

D
5 r₁

Solution

For every 10 degree Celsius rise in temperature rate is doubled. So, this option is correct.
Question 5
1 / -0

Given log 25 = 1.4

For a reaction, the rate constants at 200k and 400k temperatures are 50 x 10^-3 and 125 x 10^-2, respectively. The activation energy of the reaction (in kcals) will be

A
2763 kcals

B
3869 kcals

C
22.98 kJ

D
22.98 kcals

Solution

This is the correct answer on the basis of Arrhenius equation.
Question 6
1 / -0

P₂O₅ P₂O₃+ O₂
Given that: - d[P₂O₅]/dt = k₁[P₂O₅]
+ d[P₂O₃]/dt = k₂[P₂O₅]
+ d[ O₂]/dt = k₃[P₂O₅]
[Neglect the feasibility of the reaction]

The relation between k₁, k₂and k₃ is

A
k₁= k₂= k₃

B
2k₁= k₂= k₃

C
3k₁= k₂= k₃

D
k₁= k₂= 2k₃

Solution

Since the number of moles are the same for all the reactants and products, all rate constants will be the same.
Question 7
1 / -0

Consider : a = initial concentration
a-x = current concentration at a time t

A straight line is drawn taking 1/a-x on y-axis and time on x-axis with a slope equal to rate constant with an intercept 1/a on y-axis. What is the order of the reaction?

A
1

B
2

C
0

D
3

Solution

For the zero order reaction, a straight line is obtained on taking concentration on y-axis vs time on x-axis.
Question 8
1 / -0

If the concentration is made x times for an n^th order reaction, then the change in rate will be

A
x times the initial rate

B
xⁿtimes the initial rate

C
x² times the initial rate

D
x^n-1times the initial rate

Solution

This change in concentration will be for an n^thorder reaction surely. This is based on Ostwald dilution law.
Question 9
1 / -0

In a certain first order reaction after 10 minutes, the volume of the product formed is 90 litres and after completion of reaction, the product formed is 100 litres. The value of rate constant k (mole/lit.sec) is

A
0.2303

B
23.03

C
2.303

D
10.0

Solution

Substitute the values in first order integral equation. The answer on simplification comes out to be 0.2303, which is correct.
Question 10
1 / -0

The half-life periods of four reactions A, B, C and D are 25 sec, 1.5 minutes, 200 sec and 0.2 minutes, respectively. The fastest reaction is

A
A

B
B

C
C

D
D

Solution

Lower the half life, faster is the reaction.
Question 11
1 / -0

What is the rate of a first order reaction whose half life is 10 seconds and initial concentration is 10 mole/lit?

A
0.0693 moles/lit.sec

B
0.00693 lit/mole.sec

C
0.00693 sec^-1

D
0.00693 moles/lit.sec

Solution

From the half life given, first calculate rate constant.
Then substitute the value in integral rate equation to find the value of the rate of reaction.
Question 12
1 / -0

C₁₂H₂₂O₁₁+ H₂ O C₆H₁₂O₆+ C₆H₁₂O₆

The order of the above reaction is

A
1

B
0

C
2

D
4

Solution

This reaction is popularly known as pseudo unimolecular first order reaction. The rate of the reaction is dependent on concentration of sucrose only. So, this is the first order reaction.
Question 13
1 / -0

SO₂ + O₂ 2SO₃

If the rate of disappearance of sulphur dioxide is 0.01 moles/lit, the rate of appearance of sulphur trioxide will be

A
0.005 moles/lit

B
0.01 moles/lit

C
0.02 moles/lit

D
0.002 moles/lit

Solution

Rate of formation of SO₃ is double the rate of disappearance of SO₂according to the given chemical reaction.
Question 14
1 / -0

H₂+ I₂ 2HI
Given that - d[H₂]/dt = k₁[H₂][I₂]
- d[I₂]/dt = k₂[H₂][I₂]
+1/2 d[HI]/dt = k₃[H₂][I₂]

What is the relation between k_1,k₂and k₃?

A
k₁=k₂= 1/2k₃

B
k₁= k₂= 2k₃

C
2k₁=k₂= k₃

D
2k₁= 2k₂= k₃

Solution

This exactly follows the rate expression for the given chemical reaction.

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Chemical Kinetics Test - 3

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Chemical Kinetics Test - 3

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Question 1

1

0

2

4

Solution

Question 2

logarithm of rate constant (y-axis) against reciprocal of absolute temperature (x-axis)

logarithm of rate constant (y-axis) against time (x-axis)

rate constant (y-axis) against reciprocal of time (x- axis)

logarithm of rate constant (y-axis) against logarithm of reciprocal of absolute temperature (x-axis)

Solution

Question 3

1/k

1/k[A]3

1/k[A]2

1/k[A]

Solution

Question 4

r1/2

2 r1

10 r1

5 r1

Solution

Question 5

2763 kcals

3869 kcals

22.98 kJ

22.98 kcals

Solution

Question 6

k1 = k2 = k3

2k1 = k2 = k3

3k1 = k2 = k3

k1 = k2 = 2k3

Solution

Question 7

1

2

0

3

Solution

Question 8

x times the initial rate

xn times the initial rate

x2 times the initial rate

xn-1 times the initial rate

Solution

Question 9

0.2303

23.03

2.303

10.0

Solution

Question 10

A

B

C

D

Solution

Question 11

0.0693 moles/lit.sec

0.00693 lit/mole.sec

0.00693 sec-1

0.00693 moles/lit.sec

Solution

Question 12

1

0

2

4

Solution

1/k[A]³

1/k[A]²

r₁/2

2 r₁

10 r₁

5 r₁

k₁= k₂= k₃

2k₁= k₂= k₃

3k₁= k₂= k₃

k₁= k₂= 2k₃

xⁿtimes the initial rate

x² times the initial rate

x^n-1times the initial rate

0.00693 sec^-1

k₁=k₂= 1/2k₃

k₁= k₂= 2k₃

2k₁=k₂= k₃

2k₁= 2k₂= k₃