Thermodynamics Test

Result

Thermodynamics Test - 19

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Weekly Quiz Competition

Question 1
4 / -1

For the reactions A → B; ∆H = + 24 kJ / mol and B → C ; ∆H = -18 kJ / mol, the decreasing order of enthalpy of A, B, C follows the order

A
A ,B ,C

B
B ,C , A

C
C, B, A

D
C, A ,B

Solution
Question 2
4 / -1

The heat of combustion of solid benzoic acid at constant volume is -321.30 kJ at 27° C. The heat of combustion at constant pressure is

A
- 321.30 - 300 R

B
- 321.30 + 300 R

C
- 321.30 - 150 R

D
- 321.30 + 900 R
Question 3
4 / -1

A

B

C

D
Question 4
4 / -1

If ∆E is the heat of reaction for
C₂ H₅ OH( l ) + 3O₂ (g) →2CO₂(g) + 3H₂O( l )
at constant volume, then ∆H (heat of reaction at constant pressure), at constant temperature is

A
∆H = ∆E + RT

B
∆H = ∆E - RT

C
∆H = ∆E - 2RT

D
∆H = ∆E + 2RT

Solution
Question 5
4 / -1

The heat of formations for CO₂ (g), H₂ O( l ) and CH₄ (g) are - 400 kJ mol^-1 , - 280 kJ mol^-1 and -70 kJ mol^-1 respectively. The heat of combustion of CH₄ in kJ mol^-1 is

A
890

B
- 160

C
- 890

D
- 90
Question 6
4 / -1

The bond dissociation energies of H₂ , Cl₂ and HCl are 104, 58 and 103 kcal mol^-1 respectively. The enthalpy of formation of HCl would be

A
- 22 kcal mol^-1

B
- 44 kcal mol^-1

C
+ 44 kcal mol^-1

D
+ 22 kcal mol^-1
Question 7
4 / -1

∆H for the reaction,
C(graphite) + 2H₂ (g) → CH₄ (g) at 298 K and 1 atm is - 17900 cal. The ∆E for the above conversion would be

A
-17900 cal

B
17900 cal

C
17304 cal

D
- 17304 cal

Solution
Question 8
4 / -1

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

A
4.100 kJ mol^-1

B
3.7904 kJ mol^-1

C
37.904 kJ mol^-1

D
41.00 kJ mol^-1
Question 9
4 / -1

A
- 261 kJ

B
+ 103 kJ

C
+ 261 kJ

D
- 103 kJ
Question 10
4 / -1

A
286.6 kJ

B
573.2 kJ

C
- 28.66 kJ

D
zero

Solution