Thermodynamics Test

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Thermodynamics Test - 25

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Weekly Quiz Competition

Question 1
4 / -1

For a chemical reaction, ∆G will always be negative if

A
∆H and T∆S both are positive

B
∆H and T∆S both are negative

C
∆H is negative and T∆S is positive

D
∆H is positive and T∆S is negative
Question 2
4 / -1

I₂(s) ⇌ I₂(g) ; ∆H = + 40 kcal, ∆S = 80 cal. The sublimation point of I₂(s) will be

A
100°C

B
127°C

C
227°C

D
500°C
Question 3
4 / -1

A
1250 K

B
500 K

C
750 K

D
1000 K

Solution
Question 4
4 / -1

The standard free energy change of a reaction is ∆G° = -115 kJ at 298 K. Calculate the equilibrium constant K_p in log K_p [R = 8.314 JK^-1 mol^-1 ]

A
20.16

B
2.303

C
2.016

D
13.83
Question 5
4 / -1

If an endothermic reaction occurs spontaneously at constant temperature (T) and pressure (p), then which of the following is true ?

A
∆G > 0

B
∆H < 0

C
∆S > 0

D
∆S < 0
Question 6
4 / -1

A spontaneous process is one in which the system suffers

A
no energy change

B
a lowering of free energy

C
a lowering of entropy

D
an increase in internal energy
Question 7
4 / -1

When compared to ∆G° for the formation of Al₂O₃, the ∆G° for the formation of Cr₂O₃ is

A
higher

B
lower

C
same

D
unpredicted
Question 8
4 / -1

For a reaction to be spontaneous at all temperatures

A
∆G and ∆H should be negative

B
∆G and ∆H should be positive

C
∆G = ∆S = 0

D
∆H < ∆G
Question 9
4 / -1

The latent heat of fusion of ice at 0° C is 80 cal/g. Entropy change (∆S) accompanying the melting of 1 g of ice at 0° C would be (units cal/g/K)

A
273

B
8.0

C
0.9

D
0.293
Question 10
4 / -1

For the process H₂O(l) (1 bar, 273 K) ⇌ H₂O(g ) (1 bar, 273 K) the correct set of thermodynamic parameters is

A
∆G = 0, ∆S = +ve

B
∆G = 0, ∆S = -ve

C
∆G = +ve, ∆S = 0

D
∆G = -ve, ∆S = +ve

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Thermodynamics Test - 25

Result

Thermodynamics Test - 25

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SHARING IS CARING

Question 1

∆H and T∆S both are positive

∆H and T∆S both are negative

∆H is negative and T∆S is positive

∆H is positive and T∆S is negative

Question 2

100°C

127°C

227°C

500°C

Question 3

1250 K

500 K

750 K

1000 K

Solution

Question 4

20.16

2.303

2.016

13.83

Question 5

∆G > 0

∆H < 0

∆S > 0

∆S < 0

Question 6

no energy change

a lowering of free energy

a lowering of entropy

an increase in internal energy

Question 7

higher

lower

same

unpredicted

Question 8

∆G and ∆H should be negative

∆G and ∆H should be positive

∆G = ∆S = 0

∆H < ∆G

Question 9

273

8.0

0.9

0.293

Question 10

∆G = 0, ∆S = +ve

∆G = 0, ∆S = -ve

∆G = +ve, ∆S = 0

∆G = -ve, ∆S = +ve

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