Chemistry Test

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Chemistry Test - 2

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Weekly Quiz Competition

Question 1
3 / -1

From the rate expression for the following reaction, determine its order of reaction and the dimensions of the rate constant.

H₂O₂(aq)+3I⁻(aq)+2H⁺⟶2H2O(l)⁺I−₃;Rate=k[H₂O₂][I⁻]

A
Order = 2, Dimensions of k=mol−1Lsec⁻¹

B
Order = 0, Dimensions of k=mol−1Lsec⁻¹

C
Order = 1, Dimensions of k=mol−1Lsec⁻¹

D
Order = 1/2, Dimensions of k=mol−1Lsec⁻¹

Solution

Solution

Rate = k[H₂O₂] [l−]

Order of reaction =1 + 1 = 2
Question 2
3 / -1

An archaeologist unearths a bone sample and wants to know the age of the bone. Her chemist friend determines the 45.3% of the initial amount of carbon-14 is present in the bone sample. If the half-life of carbon-14 is 5730 yrs, calculate the age (in years) of the bone.

A
65 yrs

B
650 yrs

C
6546 yrs

D
650000 yrs

Solution

Solution
Question 3
3 / -1

A substance 'A' decomposes by a first order reaction starting initially with [A]=2.00 M and after 200 min, [A] becomes 0.15 M . For this reaction, t_1/2 is

A
53.49 min

B
50.49 min

C
48.45 min

D
46.45 min

Solution

Solution

The given reaction started initially with [A] =2.00 M

Time taken, (t)=200 min and [A] becomes 0.15 M

For a first order reaction, the rate constant is
Question 4
3 / -1

Arrhenius equation may not be represented as:

A

B

C

D

Solution

Solution

The Arrhenius equation gives the relations between rate constants at different temperatures as:
Question 5
3 / -1

For the first-order decomposition reaction of N2O5 , it is observed that,

Which of the following is true?

A
k=k'

B
k=2k'

C
k=k'/2

D
k=(k')²

Solution

Solution

For the 1st reaction,
Question 6
3 / -1

What should be the age of the fossil for meaningful determination of its age?

A
6 years.

B
6000 years.

C
60,000 years.

D
Can be used to calculate any age.

Solution

Solution

As the half life of C¹⁴is 5760 years, so a 6 year old fossil’s age can’t be determined.

Further this technique cannot be used to date objects older than 30,000 years.

After this length of time the radioactivity is too low to be measured.
Question 7
3 / -1

Two substances A (t_1/2=5 min) and B (t_1/2=15 min) are taken in such a way that initially [A]=4[B] . The time after which both the concentration will be equal is : (Assume that reaction is first order)

A
5 min

B
15 min

C
20 min

D
Concentration can never be equal.

Solution

Solution

C_t=C₀e^−Kt

According to the question,
Question 8
3 / -1

In the start of summer, a given sample of milk turns sour at room temperature (27^oC) in 48 hour . In a refrigerator at 2oC , milk can be stored three times before it sours. For souring of milk, At 2^oC , the reaction is three times slower than at 27^oC . The activation energy of the souring of milk is (in kJ mol⁻¹) .

A
30.210

B
30.146

C
30.0

D
35.126

Solution

Solution

At 2^oC (275 K) , the reaction is three times slower than at 27^oC (300 K) . This implies that for souring of milk.
Question 9
3 / -1

A + B → Product

If concentration of A is doubled, rate increases 4 times. If concentration of A and B are doubled, rate increases 8 times. The differential rate equation of the reaction will be

A

B

C

D

Solution

Solution

Let the order with respect to A and B is x and y respectively.

Hence,

Rate r=[A]^x[B]^y ...(i)

On doubling the concentration of A, rate increases 4 times,

4r=[2A]^x[B]^y ... (ii)

From Eqs. (i) and (ii)
Question 10
3 / -1

The rate constant for the reaction, 2N₂O₅→4NO₂+O₂ is 2×10^–5s^–1. If rate of reaction is 1.4×10^–5mol L^–1s^–1 , what will be the concentration of N₂O₅ in mol L^–1 ?

A
0.8

B
0.7

C
1.2

D
1

Solution

Solution

Rate = k[N₂O₅] (first order as unit of rate constant is s^–1)