Thermodynamics Test

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Thermodynamics Test - 54

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Weekly Quiz Competition

Question 1
4 / -1

An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If Ti is the initial temperature and Tf is the final temperature, which of the following statements is correct ?

A
T_f > T_i for reversible process but T_f = T_i for irreversible process

B
(T_f)_rev = (T_f)_irev

C
T_f = T_i for both reversible and irreversible processes

D
(T_f)_rev > (T_f)_irev
Question 2
4 / -1

Identify the correct statement regarding a spontaneous process :

A
Exothermic processes are always spontaneous

B
Lowering of energy in the reaction process is the only criterion for spontaneity

C
For a spontaneous process in an isolated system, the change in entropy is positive

D
Endothermic processes are never spontaneous
Question 3
4 / -1

In conversion of lime–stone to lime, CaCO₃(s) → CaO(s) + CO₂(g) the values of ∆H^o and ∆S^o are +179.1 kJ mol^–1 and 160.2 J/K respectively at 298 K and 1 bar. Assuming that ∆H° and ∆S° do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is :

A
845 K

B
1118 K

C
1008 K

D
1200 K

Solution
Question 4
4 / -1

For a reversible process at T = 300 K, the volume is increased from V_i = 1 L to V_f = 10 L. Calculate ∆H if the process is isothermal

A
11.47 kJ

B
4.98 kJ

C
0

D
– 11.47 kJ
Question 5
4 / -1

If at 298 K the bond energies of C – H, C – C, C = C and H–H bonds are respectively 414, 347, 615 and 435 kJ mol^–1, the value of enthalpy change for the reaction ; H₂C = CH₂(g) + H₂(g) → H₃C – CH₃(g) at 298 K will be –

A
+125 kJ

B
—125 kJ

C
+250 kJ

D
— 250 kJ
Question 6
4 / -1

Considering entropy(s) as thermodynamic parameter, the criterion for the spontaneity of any process is :

A
∆S_system + ∆S_surroundings > 0

B
∆S_system – ∆S_surroundings > 0

C
∆S_system > 0 only

D
∆S_surroundings > 0 only
Question 7
4 / -1

Assuming that water vapour is an ideal gas, the internal energy change (∆U) when 1 mol of water is vapourisedat 1 bar pressure and 100°C, (Given : Molar enthalpy of vapourization of water at 1 bar and 373 K = 41 kJ mol^–1 and R = 8.3 J mol^–1 K–1) will be :

A
37.904 kJ mol^–1

B
41.00 kJ mol^–1

C
4.100 kJ mol^–1

D
3.7904 mol^–1

Solution
Question 8
4 / -1

The standard enthalpy of formation (DHf°) at 398 K for methane, CH₄(g) is 74.8 kJ mol^–1. The additional information required to determine the average energy for C – H bond formation would be.

A
the dissociation energy of H₂ and enthalpy of sublimation of carbon

B
latent heat of vapourisation of methane

C
the first four ionization energies of carbon and electron gain enthalpy of hydrogen

D
the dissociation energy of hydrogen molecule, H₂

Solution
Question 9
4 / -1

Standard entropy of X₂, Y₂ and XY₃ are 60, 40 and 50 JK^–1 mol^–1, respectively. For the reaction,1/2 X₂ + 3/2 Y₂ → XY₃ DH = – 30 kJ. To be at equilibrium the temperature will be :

A
500 K

B
750 K

C
1000 K

D
1250 K

Solution
Question 10
4 / -1

On the basis of the following thermo chemical data : (∆_ƒGºH+ (aq) = 0) H₂O(l) → H+ (aq) + OH– (aq.) ; ∆H = 57.32 kJ H₂(g) + O₂(g) → H₂O(l); ∆H = –286.20 kJ The value of enthalpy of formation of OH– ion at 25ºC is :

A
–228.88 kJ

B
+228.88 kJ

C
–343.52 kJ

D
–22.88 kJ

Solution

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Thermodynamics Test - 54

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Thermodynamics Test - 54

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Question 1

Tf > Ti for reversible process but Tf = Ti for irreversible process

(Tf)rev = (Tf)irev

Tf = Ti for both reversible and irreversible processes

(Tf)rev > (Tf)irev

Question 2

Exothermic processes are always spontaneous

Lowering of energy in the reaction process is the only criterion for spontaneity

For a spontaneous process in an isolated system, the change in entropy is positive

Endothermic processes are never spontaneous

Question 3

845 K

1118 K

1008 K

1200 K

Solution

Question 4

11.47 kJ

4.98 kJ

0

– 11.47 kJ

Question 5

+125 kJ

—125 kJ

+250 kJ

— 250 kJ

Question 6

∆Ssystem + ∆Ssurroundings > 0

∆Ssystem – ∆Ssurroundings > 0

∆Ssystem > 0 only

∆Ssurroundings > 0 only

Question 7

37.904 kJ mol–1

41.00 kJ mol–1

4.100 kJ mol–1

3.7904 mol–1

Solution

Question 8

the dissociation energy of H2 and enthalpy of sublimation of carbon

latent heat of vapourisation of methane

the first four ionization energies of carbon and electron gain enthalpy of hydrogen

the dissociation energy of hydrogen molecule, H2

Solution

Question 9

500 K

750 K

1000 K

1250 K

Solution

Question 10

–228.88 kJ

+228.88 kJ

–343.52 kJ

–22.88 kJ

Solution

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T_f > T_i for reversible process but T_f = T_i for irreversible process

(T_f)_rev = (T_f)_irev

T_f = T_i for both reversible and irreversible processes

(T_f)_rev > (T_f)_irev

∆S_system + ∆S_surroundings > 0

∆S_system – ∆S_surroundings > 0

∆S_system > 0 only

∆S_surroundings > 0 only

37.904 kJ mol^–1

41.00 kJ mol^–1

4.100 kJ mol^–1

3.7904 mol^–1

the dissociation energy of H₂ and enthalpy of sublimation of carbon

the dissociation energy of hydrogen molecule, H₂