Thermodynamics Test

Result

Thermodynamics Test - 20

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
4 / -1

For an ideal gas, the heat of reaction at constant pressure and constant volume are related as

A
H + E = pV

B
E = H + p∆V

C
q_p = q_v + ∆nRT

D
None of these
Question 2
4 / -1

Consider the reaction, N₂ + 3H₂ → 2NH₃ carried out at constant temperature and pressure. If ∆H and ∆U are the enthalpy and internal energy changes for the reaction, which of the following expressions is true?

A
∆H > ∆U

B
∆H < ∆U

C
∆H = ∆U

D
∆H = 0
Question 3
4 / -1

The enthalpies of combustion of carbon and carbon monoxide are -393.5 and- 283 kJ mol^-1 respectively. The enthalpy of formation of carbon monoxide per mole is

A
110.5 kJ

B
676.5 kJ

C
- 676.5 kJ

D
- 110.5 kJ
Question 4
4 / -1

Calculate enthalpy for formation of ethylene from the following data

A
54.1 kJ

B
44.8 kJ

C
51.4 kJ

D
48.4 kJ
Question 5
4 / -1

(∆H - ∆U) for the formation of carbon monoxide (CO) from its elements at 298 K is (R = 8.314 JK^-1 mol^-1 )

A
-1238.78 J mol^-1

B
1238.78 J mol^-1

C
-2477.57 mol^-1

D
2477.57 J mol^-1
Question 6
4 / -1

If heat of neutralisation of HCN with NaOH is -12.1 kJ and the heat of neutralisation of HCl by NaOH is -55.9 kJ/mol, then the energy of dissociation of HCN is

A
43.8 kJ

B
-43.8 kJ

C
-68 kJ

D
68 kJ
Question 7
4 / -1

Heat of formation of SO₂ is - 298kJ. What is the heat of combustion of 4 g of S ?

A
+ 37 kJ

B
-37.25 kJ

C
+298 kJ

D
18.6 kJ
Question 8
4 / -1

The heats of combustion of carbon monoxide at constant pressure and at constant volume at 27° C will differ from one another by

A
27 cal

B
54 cal

C
300 cal

D
600 cal
Question 9
4 / -1

For the reaction, A (g)+ 2B (g) → 2C (g)+ 3D (g) the change of enthalpy at 27° C is 19 kcal. The value of ∆E is

A
21.2 kcal

B
17.8 kcal

C
18.4 kcal

D
20.6 kcal
Question 10
4 / -1

In an isochoric process, ∆H for a system is equal to

A
p . ∆V

B
pV

C
E + p . ∆V

D
∆E

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Thermodynamics Test - 20

Result

Thermodynamics Test - 20

Score

-

Rank

-

SHARING IS CARING

Question 1

H + E = pV

E = H + p∆V

qp = qv + ∆nRT

None of these

Question 2

∆H > ∆U

∆H < ∆U

∆H = ∆U

∆H = 0

Question 3

110.5 kJ

676.5 kJ

- 676.5 kJ

- 110.5 kJ

Question 4

54.1 kJ

44.8 kJ

51.4 kJ

48.4 kJ

Question 5

-1238.78 J mol-1

1238.78 J mol-1

-2477.57 mol-1

2477.57 J mol-1

Question 6

43.8 kJ

-43.8 kJ

-68 kJ

68 kJ

Question 7

+ 37 kJ

-37.25 kJ

+298 kJ

18.6 kJ

Question 8

27 cal

54 cal

300 cal

600 cal

Question 9

21.2 kcal

17.8 kcal

18.4 kcal

20.6 kcal

Question 10

p . ∆V

pV

E + p . ∆V

∆E

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

q_p = q_v + ∆nRT

-1238.78 J mol^-1

1238.78 J mol^-1

-2477.57 mol^-1

2477.57 J mol^-1