Chemistry Test

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Chemistry Test - 4

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Question 1
1 / -0

Directions: In the following question, a statement of assertion is given, followed by a corresponding statement of the reason below it. Choose the correct option.

Assertion: Cu²⁺ and Cd²⁺are separated by first adding KCN solution and then, passing H₂S gas.

Reason: KCN reduces Cu²⁺ to Cu⁺and forms a complex with it.

A
Both assertion and reason are true and reason is the correct explanation of assertion.

B
Both assertion and reason are true, but reason is not the correct explanation of assertion.

C
Assertion is true, but reason is false.

D
Both assertion and the reason are false.

Solution

Both assertion and reason are true, but the reason is not a correct explanation for the assertion.

Cu⁺² makes a more stable complex with CN^- which is soluble and therefore the solution, now contains, the complex Cu⁺² ion and the free Cd⁺² ion. Hence only the sulphide of Cadmium will precipitate out on the passage of H₂S gas.
Question 2
1 / -0

Directions: In the following question, a statement of Assertion is given; followed by a corresponding statement of Reason just below it. Out of the statements mark the correct answer as:

(a) If both assertion and reason are true and the reason is the correct explanation of the assertion.

(b) If both assertion and reason are true, but the reason is not the correct explanation of the assertion.

(c) If assertion is true statement, but reason is false.

(d) If both assertion and reason are false.

Assertion: According to transition state theory, for the formation of an activated complex; one of the vibrational degree of freedom is converted into a translational degree of freedom.

Reason: Energy of the activated complex is higher than the energy of reactant molecules.

A
(a)

B
(b)

C
(c)

D
(d)

Solution

Transition state theory (TST) explains the reaction rates of elementary chemical reactions. The theory assumes a special type of chemical equilibrium (quasi-equilibrium) between reactants and activated transition state complexes. TST is used primarily to understand qualitatively how chemical reactions take place. In other words, one of the vibrational degrees of freedom is converted into translational degree of freedom.

An activated complex acts as an intermediary between the reactants and the products of the reaction. The energy of the activated complex is higher than that of reactants or the products, and the state is temporary. If there is not sufficient energy to sustain the chemical reaction, the activated complex can reform into the reactants in a backward reaction. With proper energy, though, the activated complex forms the products in a forward reaction.
Question 3
1 / -0

For a hypothetical reaction 2A + 3B→2C, the rate law expression is = dx/dt k[A] [B]². The order of the reaction is

A
1

B
2

C
2.5

D
3

Solution

In chemical kinetics, the order of reaction with respect to a given substance (such as reactant, catalyst or product) is defined as the index, or exponent, to which its concentration term in the rate equation is raised. For the given expression, according to rate law, order of reaction is: 2 + 1 = 3.
Question 4
1 / -0

The plot of ln C_{0 /}C_t against t is a straight line; the order of the reaction is expected to be:

A
second order

B
first order

C
zero order

D
pseudo first order

Solution
Question 5
1 / -0

A reaction was found to be in second order with respect to the concentration of carbon monoxide. If the concentration of carbon monoxide is doubled with everything else kept the same, the rate of reaction will

A
be doubled

B
remain unchanged

C
be tripled

D
ncrease by a factor of 4

Solution
Question 6
1 / -0

Unit of rate constant of zero order reaction is

A
time^-1

B
mol lit^-1 sec^-1

C
liter mol^-1 sec^-1

D
liter² mol^-2sec^-1

Solution

According to rate law for zero order reaction,

Rate = K [A]⁰

= K = Mol L^-1s^-1

Thus, unit of rate constant is: Mol L^-1s^-1
Question 7
1 / -0

The gas-phase reaction of nitric oxide and bromine yields nitrosyl bromide as 2NO(g) + Br₂(g) →2NOBr(g). What is the order of the reaction?

A
1

B
2

C
3

D
4

Solution

Rate law is ,

rate=k[NO]²[Br₂]

hence, wrt NO, order=2

wrt Br₂, order=1

∴overall order = 2+1=3
Question 8
1 / -0

For As₂S₃ sol, the maximum coagulation power is of

A
cl^-

B
SO₄^-2

C
PO₄^-3

D
Al⁺³

Solution

As₂S₃ is a negatively charged sol.

Hence, the cation of the electrolyte will be the active ion and option (4) is correct.
Question 9
1 / -0

Bredig's method can not be used for the preparation of collodial sol of

A
chromium

B
sodium

C
silver

D
gold

Solution

Bredig's method cannot be used for the preparation of colloidal sol of sodium as it does not give satisfactory results.
Question 10
1 / -0

If C₆H₆(I) + 15/2 O₂(g)→3H₂O(I) + 6CO₂(g); ΔH= – 3264.6 kJ mol^–1, then the energy obtained by burning 3.9 g of benzene in the air is

A
163.23 kJ

B
326.4 kJ

C
32.64 kJ

D
16.32 kJ

Solution

78 g of C₆H₆ gives heat

= 3264.6 kJ

∴3.9 g of C₆H₆ will give heat

= (3264.6 / 78) x 3.9= 163.23 kJ

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Chemistry Test - 4

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Chemistry Test - 4

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Question 1

Both assertion and reason are true and reason is the correct explanation of assertion.

Both assertion and reason are true, but reason is not the correct explanation of assertion.

Assertion is true, but reason is false.

Both assertion and the reason are false.

Solution

Question 2

(a)

(b)

(c)

(d)

Solution

Question 3

1

2

2.5

3

Solution

Question 4

second order

first order

zero order

pseudo first order

Solution

Question 5

be doubled

remain unchanged

be tripled

ncrease by a factor of 4

Solution

Question 6

time-1

mol lit-1 sec-1

liter mol-1 sec-1

liter2 mol-2sec-1

Solution

Question 7

1

2

3

4

Solution

Question 8

cl-

SO4-2

PO4-3

Al+3

Solution

Question 9

chromium

sodium

silver

gold

Solution

Question 10

163.23 kJ

326.4 kJ

32.64 kJ

16.32 kJ

Solution

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time^-1

mol lit^-1 sec^-1

liter mol^-1 sec^-1

liter² mol^-2sec^-1

cl^-

SO₄^-2

PO₄^-3

Al⁺³