Chemistry Test

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Chemistry Test - 11

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Weekly Quiz Competition

Question 1
1 / -0

The correct order of bond energies of the following bonds is:

(I) C(sp³)-C(sp³), (II) C(sp²)-C(sp²), (III) C(sp)-C(sp)

A
I < II < III

B
I > II > III

C
II > III > I

D
III > I > II

Solution

I < II < III
Question 2
1 / -0

A catalyst is added to a reaction to increase the rate of reaction. Which of the following statements is correct?

A
A catalyst increases the activation energy between reactants and products and hence, lowers the potential energy barrier.

B
A catalyst provides an alternate pathway or reaction mechanism.

C
A catalyst alters $∆$ G, Gibbs energy of reaction.

D
A catalyst changes the equilibrium constant of a reaction.

Solution

A catalyst provides an alternate pathway or reaction mechanism.

Question 3

1 / -0

Directions: Match the order of the reaction from column - I with the reaction characteristics from column - II in the following table.

Column - I	Column - II
(A) Zero-order reaction	(a) High activation energy
(B) Pseudo unimolecular reaction	(b) Decreases activation energy
(C) Slow reaction	(c) Decomposition of ammonia on tungsten
(D) Positive catalyst	(d) Hydrolysis of cane sugar

(A) - (c), (B) - (d), (C) - (a), (D) - (b)

(A) - (d), (B) - (a), (C) - (c), (D) - (b)

(A) - (c), (B) - (b), (C) - (a), (D) - (d)

(A) - (a), (B) - (d), (C) - (c), (D) - (b)

Solution

(A) - (c), (B) - (d), (C) - (a), (D) - (b)

Question 4
1 / -0

Directions: The following question has four choices out of which ONLY ONE is correct.

The experimental data for the reaction 2A + B₂ $\to$ 2AB is given in the table below:

Experiment [A] [B₂] Rate (mole s ^-1)

1 0.50 0.50 1.6 ×10^–4

2 0.50 1.00 3.2 ×10^–4

3 1.00 1.00 3.2 ×10^–4

The rate equation for the above data is

A
rate = k[B₂]

B
rate = k[B₂]²

C
rate = k[A]²[B]²

D
rate = k[A]²[B]

Solution

rate = k[B₂]
Question 5
1 / -0

We wish to see inside an atom. Assume the atom to have a diameter of 100 pm. This means that one must be able to resolve a width of say 10 pm. If an electron microscope is used, the energy required should be

A
1.5 keV

B
15 keV

C
150 keV

D
1500 keV

Solution

Correct option is 15 keV.
Question 6
1 / -0

The half-life of 131I is 8 days. If we have a sample of 131I at time t = 0, we can assert that

A
no nucleus will decay before t = 4 days

B
no nucleus will decay before t = 8 days

C
all nuclei will decay before t = 16 days

D
a given nucleus may decay any time after t = 0

Solution

a given nucleus may decay any time after t = 0
Question 7
1 / -0

A radioactive nuclide can decay simultaneously by two different processes, which have individual decay constants $\lambda_{1}$ and $\lambda_{2}$. The effective decay constant of the nuclide ( $\lambda$ ) is given by

A
$\lambda=\sqrt{\lambda_{1} \lambda_{2}}$

B
$\frac{1}{\lambda}=\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}$

C
$\lambda=\frac{1}{2}\left(\lambda_{1}+\lambda_{2}\right)$

D
$\lambda=\lambda_{1}+\lambda_{2}$

Solution

$\lambda=\lambda_{1}+\lambda_{2}$
Question 8
1 / -0

The binding energies for nuclei ₁H¹, ₂He⁴, ₂₆Fe⁵⁶ and ₉₂U²³⁵ are 2.22, 28.3, 492 and 17.86 MeV, respectively. The most stable nucleus is

A
₁H¹

B
₂He⁴

C
₂₆Fe⁵⁶

D
₉₂U²³⁵

Solution

The most stable nucleus is₂₆Fe⁵⁶
Question 9
1 / -0

The binding energy per nucleon of ₁H² is 1.1 MeV and binding energy per nucleon of ₂He⁴ is 7 MeV. The energy released when two ₁H² combine to form ₂He⁴ is

A
6.8 MeV

B
30.2 MeV

C
23.6 MeV

D
None of these

Solution

23.6 MeV
Question 10
1 / -0

For a certain radioactive substance, it is observed that after 4 hours, only 6.25% of the original sample is left undecayed. Which of the following statements is incorrect?

A
The half life of the sample is 1 hr.

B
The mean life of the sample is 1/2 hr.

C
The decay constant of the sample is (ln 2) hr^-1.

D
After further 4 hours, the amount of the substance leftover would be only 0.39% of the original amount.

Solution

Correct Answer : The mean life of the sample is 1/2 hr.

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Experiment	[A]	[B₂]	Rate (mole s ^-1)
1	0.50	0.50	1.6 ×10^–4
2	0.50	1.00	3.2 ×10^–4
3	1.00	1.00	3.2 ×10^–4

Chemistry Test - 11

Result

Chemistry Test - 11

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SHARING IS CARING

Question 1

I < II < III

I > II > III

II > III > I

III > I > II

Solution

Question 2

A catalyst increases the activation energy between reactants and products and hence, lowers the potential energy barrier.

A catalyst provides an alternate pathway or reaction mechanism.

A catalyst alters ∆G, Gibbs energy of reaction.

A catalyst changes the equilibrium constant of a reaction.

Solution

Question 3

(A) - (c), (B) - (d), (C) - (a), (D) - (b)

(A) - (d), (B) - (a), (C) - (c), (D) - (b)

(A) - (c), (B) - (b), (C) - (a), (D) - (d)

(A) - (a), (B) - (d), (C) - (c), (D) - (b)

Solution

Question 4

rate = k[B2]

rate = k[B2]2

rate = k[A]2[B]2

rate = k[A]2[B]

Solution

Question 5

1.5 keV

15 keV

150 keV

1500 keV

Solution

Question 6

no nucleus will decay before t = 4 days

no nucleus will decay before t = 8 days

all nuclei will decay before t = 16 days

a given nucleus may decay any time after t = 0

Solution

Question 7

\(\lambda=\sqrt{\lambda_{1} \lambda_{2}}\)

\(\frac{1}{\lambda}=\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}\)

\(\lambda=\frac{1}{2}\left(\lambda_{1}+\lambda_{2}\right)\)

\(\lambda=\lambda_{1}+\lambda_{2}\)

Solution

Question 8

1H1

2He4

26Fe56

92U235

Solution

Question 9

6.8 MeV

30.2 MeV

23.6 MeV

None of these

Solution

Question 10

The half life of the sample is 1 hr.

The mean life of the sample is 1/2 hr.

The decay constant of the sample is (ln 2) hr-1.

After further 4 hours, the amount of the substance leftover would be only 0.39% of the original amount.

Solution

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A catalyst alters $∆$ G, Gibbs energy of reaction.

rate = k[B₂]

rate = k[B₂]²

rate = k[A]²[B]²

rate = k[A]²[B]

₁H¹

₂He⁴

₂₆Fe⁵⁶

₉₂U²³⁵

The decay constant of the sample is (ln 2) hr^-1.