Thermodynamics Test

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Thermodynamics Test - 26

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Weekly Quiz Competition

Question 1
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Direction (Q. Nos. 1- 8) This section contains 8 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.
Q. For the given reaction, = - 1.3818 kcal at 300 K. Thus equilibrium constant is

A
1

B
0

C
0.1

D
100

Solution

Since the reaction is not reversible, equilibrium constant is zero.
Question 2
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ΔH_vap = 30 kJ mol^-1 and ΔS_vap = 75 J mol^-1 K^-1. Thus, temperature of vapour at one atmosphere is
[IIT JEE 2004]

A
400 K

B
350 K

C
298 K

D
250 K

Solution

∆H_Vap = 30kJ/mol and ∆S = 75 J/K
∆G = ∆H - T∆S
At eqm, ∆G = 0
Therefore, ∆H = T∆S
Or T = 30×10³/75 = 400 K
Question 3
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Which reaction, with the following values of ΔH and ΔS at 400 K is spontaneous and endothermic?

A
a

B
b

C
c

D
d

Solution

∆G = ∆H - T∆S
For opt (c), ∆G = 48000 - 400(135)
= 48000 - 54000
= -6000
∆G is -ve
Therefore reaction is spontaneous.
Question 4
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Standard entropies of X₂, Y₂ and XY₃ are given below the reaction
Q. At what temperature, reaction would be in equilibrium?

A
500 K

B
750 K

C
1000 K

D
1250 K

Solution

1/2X₂ + 3/2Y₂ ⟶XY₃,
ΔH= −30 kJ
ΔS_reaction = ∑ΔS_product−∑ΔS_reactant
X₂+ 3Y₂→ 2XY₃
ΔH=−60 kJ
ΔS_reaction= 2×50−3×40−1×60 =100−120−60=−80 JK⁻¹mol⁻¹
ΔG=ΔH−TΔS=0
ΔH=TΔS
1000×(−60)=−80×T
T=750 K
Question 5
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The value of log₁₀ K for a reaction, A B is (Given, ΔH°₂₉₈ = - 54.07 kJ mol^-1;
ΔS°₂₉₈ = + 10 JK^-1 mol^-1; R = 8.314 JK^-1 mol^-1 2.303 x 8.314 x 298 = 5705)
[IITJEE2007]

A
5

B
10

C
95

D
100

Solution

∆G° = ∆H° - T∆S°
= -54070 - 298 10
= -57050
∆G° = -2.303 RT log₁₀k
-57050 = -2.303 8.314 298 log₁₀k
57050 = 5705 log₁₀k
log₁₀k = 10
Question 6
1 / -0

For the following decom position reaction,
ΔG° = 1729 J mol^-1 at 300 K , (log 5 = 0.6990]
Thus, pressure set up is

A
0.3010

B
0.6990

C
0.500

D
0.25
Question 7
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If for the cell, Zn(s) + Cu²⁺(ag) Cu(s) + Zn²⁺ (ag)entropy change ΔS° is 96.5 JK^-1 mol^-1, then temperature coefficient of the emf of a cell is

A
5 x 10^-4 VK^-1

B
1 x 10^-3 VK^-1

C
2 x 10^-3 VK^-1

D
9.65 x 10⁴ VK^-1

Solution

ΔG=ΔH−nFT(dE/dT)_P
and ΔG=ΔH−TΔS
∴ΔS/nF=(dE/dT)_P
or
96.5/2×96500=(dE/dT)_P
∴(dE_cell/dT)P
=1×10⁻³ / 2
=5×10⁻⁴VK⁻¹
Question 8
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For the reaction, at 1000° C
ΔG° = - 27 kJ mol^-1, = 0.0033 atm.
Q. Hence, ΔG at this temperature is

A
- 60.5 kJ mol^-1

B
- 87.5 kJ mol^-1

C
-33.5 kJ mol^-1

D
+ 33.5 kJ mol^-1
Question 9
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Direction (Q. No. 9) This sectionis based on statement I and Statement II. Select the correct answer from the code given below.
Q.
Statement I :Every endothermic reaction is spontaneous if TΔS > ΔH.
Statement II : Sign of ΔG is the true criterion for deciding spontaneity of a reaction.

A
Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

B
Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I

C
Statement I is correct but Statement II is incorrect

D
Statement II is correct but Statement I is incorrect
Question 10
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Direction (Q. Nos. 10) Choice the correct combination of elements and column I and coloumn II are given as option (a), (b), (c) and (d), out of which ONE option is correct.
Q. Thermodynamic equilibrium constant K is related to temperature by equation log₁₀

A
a

B
b

C
c

D
d