Thermodynamics Test

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Thermodynamics Test - 4

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Question 1
1 / -0

Change in Gibb's free energy may be expressed as

A
ΔG = ΔH - TΔS

B
ΔG = ΔH +

C
ΔH = ΔG - TΔS

D
ΔG = ΔH + T.C_P

Solution

Gibb's Helmholtz equation is ΔG = ΔH - TΔS.
The Gibbs free energy is the maximum amount of non-expansion work that can be extracted from a thermodynamically closed system (one that can exchange heat and work with its surroundings, but not matter).
Question 2
1 / -0

For the spontaneity of a cell, which of the following options is correct?

A
= 0, = 0

B
= -ve, = 0

C
= +ve, = +ve

D
= -ve, = +ve

Solution

For a spontaneous process, = -ve and for a cell = +ve.
Question 3
1 / -0

The negative value of enthalpy change when 36.5 g of HCl and 40 g of NaOH reacts during neutralisation is

A
> 6.5 k cal

B
13.6 k cal

C
> 13.6 k cal

D
108 k cal

Solution

The enthalpy of neutralization is the amount of energy released when one mole OH^- from a base is neutralized by one mole of H⁺ ions from an acid.
36.5 g of HCl = 1 mol HCl (a strong monobasic acid)
40 g of NaOH = 1 mol of NaOH (a strong monoacidic base)

The enthalpy of neutralization of a strong acid by a strong base is -57.3 kJ or - 13.6 kcal (negative sign indicates that the reaction is exothermic).
Question 4
1 / -0

When a gas undergoes adiabatic expansion, it gets cooled due to

A
loss of kinetic energy

B
fall in temperature

C
decrease in velocity

D
energy used in doing work

Solution

For the expansion of a gas under adiabatic conditions, the internal energy of the gas is utilized for the expansion work and as a result the internal energy decreases and so does the temperature of the gas.
Question 5
1 / -0

In which of the following cases does entropy decrease?

A
Crystallisation of sucrose from solution

B
Rusting of iron

C
Melting of ice

D
Vaporisation of camphor

Solution

Crystallization of sucrose from solution results in decrease in randomness. Because of which the entropy will decrease. Aqueous sucrose solution is more disordered than crystalline sucrose.On the other hand, rusting of iron, conversion of ice to water and vaporization of camphor will result in an increase in entropy.
Question 6
1 / -0

The amount of heat measured in a bomb calorimeter is

A
ΔG

B
ΔH

C
ΔE

D
PΔV

Solution

A bomb calorimeter is a type of constant-volume calorimeter used in measuring the heat of combustion of a particular reaction. Heat change measured in a closed vessel is ΔE.
Question 7
1 / -0

The heat required to raise the temperature of a body by 1 K is called

A
specific heat

B
thermal capacity

C
water equivalent

D
None of these

Solution

Thermal capacity is the amount of heat required to raise the temperature of a body by 1 K.
Specific heat = Thermal capacity × mass of the object
Water equivalent is the amount of water that would absorb the same amount of heat as the thermal capacity of the material.
Hence, option (2) is correct.
Question 8
1 / -0

For an endothermic reaction where ΔH represents the enthalpy change of the reaction in kJ/mole, the minimum value for E_a will be

A
less than ΔH

B
zero

C
more than ΔH

D
equal to ΔH

Solution

For and endothermic process the reactants are more stable than the products and activation energy for the forward reaction is always greater than the enthalpy change for the reaction.

Question 9

1 / -0

Which of the following gases has the highest heat of combustion?

Methane

Ethane

Ethene

Acetylene

Solution

The ethalpy of combustion increases with increase in the carbon content.
Among hydrocarbons having the same number of carbon atoms, saturated hydrocarbons alkanes have the maximum enthalpy of combustion.

Hydrocarbon	Enthalpy of combustion, Δ_cH°
Methane	-890.8 kJ/mol
Acetylene	-422.85 kJ/mol
Ethene	-1400 kJ/mol
Ethane	−1560.7 kJ/mol

Question 10
1 / -0

Hess's law deals with

A
change in heat of reaction

B
rate of reaction

C
equilibrium constant

D
influence of pressure on volume of a gas

Solution

Hess's law deals with heat changes during processes.
It states that:
The heat of any reaction ΔH_∘ for a specific reaction is equal to the sum of the heats of reaction for any set of reactions which in sum are equivalent to the overall reaction.
Question 11
1 / -0

A pure crystalline substance at absolute zero temperature has zero entropy is called

A
Faraday's 2^nd Law

B
Newton's 3^rd Law

C
Kohlraush Law

D
Nernst Law

Solution

Nernst 3^rd law postulated that a pure crystalline substance at absolute zero temperature has zero entropy.
Question 12
1 / -0

Biosynthesis of sucrose from glucose and fructose occurs with ΔG° of +23 kJ/mole. This reaction is called as

A
coupled reaction

B
exothermic reaction

C
biochemical reaction

D
natural reaction

Solution

Only coupled reactions are possible with ΔG° as positive.
Question 13
1 / -0

The units of entropy are

A
J/k/mol

B
J/k

C
KJ

D
J/mole

Solution

Total entropy change, which is the sum of the entropy changes of the system and the surroundings.
ΔS_Total = ΔS_Surroundings + ΔS_SystemΔS_{Surroundings =}Units of entropy change are J mol^-1K^-1.
Question 14
1 / -0

Energy of the universe is constant but the entropy of universe tends to maximize.

The above statement is implicit in which of the following law(s)?

A
Gibb's law

B
Hess's law

C
Newton's law

D
1^st and 2^nd law of thermodynamics

Solution

The first law of thermodynamics is and extension of the law of conservation of energy and is mathematically represented as:
ΔU = q + w
It implies that the energy of the universe is conserved.
The 2^nd law states that for any process,
ΔS_Total = ΔS_System + ΔS_Surrounding
For a spontaneous process,
ΔS_Total > 0
Which implies that the universe is always expanding.
Question 15
1 / -0

The absolute value of which of the following thermodynamic quantities can be determined?

A
E (Internal energy)

B
H (Enthalpy)

C
S (Entropy)

D
All of these

Solution

Internal energy and enthalpy of a system cannot be determined in absolute terms.
Although it is not practically possible to calculate the absolute entropy of a system but there is a way to theoretically approximatly calculate an absolute value for entropy using to the third law of thermodynamics.
dS=δq/T.
By the definition, a perfect crystal at 0 K would have an entropy S=0.
Hence, option (3) is correct.

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Thermodynamics Test - 4

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Thermodynamics Test - 4

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Question 1

ΔG = ΔH - TΔS

ΔG = ΔH +

ΔH = ΔG - TΔS

ΔG = ΔH + T.CP

Solution

Question 2

= 0, = 0

= -ve, = 0

= +ve, = +ve

= -ve, = +ve

Solution

Question 3

> 6.5 k cal

13.6 k cal

> 13.6 k cal

108 k cal

Solution

Question 4

loss of kinetic energy

fall in temperature

decrease in velocity

energy used in doing work

Solution

Question 5

Crystallisation of sucrose from solution

Rusting of iron

Melting of ice

Vaporisation of camphor

Solution

Question 6

ΔG

ΔH

ΔE

PΔV

Solution

Question 7

specific heat

thermal capacity

water equivalent

None of these

Solution

Question 8

less than ΔH

zero

more than ΔH

equal to ΔH

Solution

Question 9

Methane

Ethane

Ethene

Acetylene

Solution

Question 10

change in heat of reaction

rate of reaction

equilibrium constant

influence of pressure on volume of a gas

Solution

Question 11

Faraday's 2nd Law

Newton's 3rd Law

Kohlraush Law

Nernst Law

Solution

Question 12

coupled reaction

exothermic reaction

biochemical reaction

natural reaction

Solution

ΔG = ΔH + T.C_P

Faraday's 2^nd Law

Newton's 3^rd Law

1^st and 2^nd law of thermodynamics