Thermodynamics Test

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Thermodynamics Test - 3

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Weekly Quiz Competition

Question 1
1 / -0

for a forward reaction in the process A + B ⇌ C is -ve.

What is the sign of in case of a backward reaction?

A
-ve

B
+ve

C
Zero

D
None of these

Solution

A reversible reaction is spontaneous in both directions and hene, is -ve for both the forward as well as backward reaction.
Question 2
1 / -0

For the reaction
(g) (g)
The signs of and are

A
= –ve, ₌ +ve

B
= +ve, = +ve

C
= –ve, ^{= –ve}

D
= 0, = +ve

Solution

= –ve
For formation of a molecule,
= –ve (Decrease in entropy)
Question 3
1 / -0

For the reaction

C₃H₈ (g) + 5O₂ (g) 3CO₂ (g) + 4H₂O ()

At constant temperature, – is

A
– RT

B
+ RT

C
– 3RT

D
+ 3RT

Solution

For the reaction

C₃H₈ (g) + 5O₂ (g) 3CO₂ (g) + 4H₂O (l)
Question 4
1 / -0

The of CH₄ is –74.8 kJ/mol. The additional information required for determining the average C – H bond energy is

A
dissociation energy of H₂ and _sub of C

B
latent heat of vaporisation of CH₄

C
1^st four ionisation energies of C and electron gain enthalpy of H

D
dissociation energy of H₂ molecule

Solution

Dissociation energy of H₂ and _sub of C are required.
Question 5
1 / -0

If 1 mole of NH₃ and 1 mole of are mixed in a closed container to form (g), which of the following is correct?

A
>

B
=

C
<

D
– = 1.5

Solution

<
= + RT
Question 6
1 / -0

The amount of heat evolved when 500 cm³ of 0.1 M HCl is mixed with 200 ml of 0.2 M NaOH is

A
1.292 kJ

B
2.292 kJ

C
3.35 kJ

D
0.092 kJ

Solution

500 cm³ of 0.1 M HCl = = 0.05 mol

200 cm³ of 0.2 M NaOH = = 0.04 mol
Limiting reagent is 0.04 mol NaOH.
Heat released = 0.04 x 57.3 = 2.292 kJ
Question 7
1 / -0

If = 30 kJ/mol and = 75 J/K mol, then the temperature of vapours at 1 atm pressure is

A
400 K

B
350 K

C
298 K

D
250 K

Solution

T = = 400 K
Question 8
1 / -0

A reversible chemical reaction (g) ⇌ 2NO (g) + (g) is carried at 400 K with = 77.2 kJ/mol
= 122 J/K/mol

Which of the following is the correct value (x) for of the reaction?

A
28.4 kJ/mol

B
27.2 kJ/mol

C
25.8 kJ/mol

D
25.6 kJ/mol

Solution

= – = 77.2 – = 28.4 kJ/mol
Question 9
1 / -0

A reversible chemical reaction (g) ⇌ 2NO (g) + (g) is carried at 400 K with = 77.2 kJ/mol
= 122 J/K/mol
Using the value of x (from the previous question 8), the correct value of k (equilibrium constant) of the reaction is

A
2.96 10^–4

B
1.96 10^–4

C
3.96 10^–4

D
1.16 10^–4

Solution

= – RT log k
k = 1.96 10^–4
Question 10
1 / -0

Thermochemical equation for the gaseous and solid rocket fuels are

H₂ (g) + O₂ (g) H₂O () = – 286 kJ
+ O₂ (s) = – 1667.8 kJ

If equal masses of H₂ and are taken, the better rocket fuel is

A
H₂ (g)

B

C
both are equally efficient

D
can't be used as fuel

Solution

H₂ produces more heat/gm of H₂ used.
Question 11
1 / -0

Thermochemical equations for the gaseous and solid rocket fuels are
H₂ (g) + O₂ (g) H₂O () = – 286 kJ
+ O₂ (s) = – 1667.8 kJ

For the reaction
O₃ (s) (g) + O₂ (g)
The value of is _______ kJ.

A
+ 1667.8

B
– 1667.8

C
+ 3335.6

D
– 3335.6

Solution

for a reverse process is equal with opposite sign.
Question 12
1 / -0

Thermochemical equations for the combustion of some important fuels are
H₂(g) + O₂ (g) H₂O () = – 286.0 kJ
C (g) + O₂ (g) CO₂ (g) = – 393.5 kJ
C₆H₆ () + O₂ (g) 6CO₂ + 3H₂O () = – 3267.6 kJ
C₂H₅OH + 3O₂ 2CO₂ + 3H₂O = – 1368.0 kJ

The amount of heat change, when 1 mole of C₂H₅OH is formed from its constituents, is

A
+ 177.0 kJ

B
– 277.0 kJ

C
– 335 kJ

D
+ 422.45 kJ

Solution

of C₂H₅OH = – 277.0 kJ
Question 13
1 / -0

Thermochemical equations for the combustion of some important fuels are
H₂(g) + O₂ (g) H₂O () = – 286.0 kJ
C (g) + O₂ (g) CO₂ (g) = – 393.5 kJ
C₆H₆ () + O₂ (g) 6CO₂ + 3H₂O () = – 3267.6 kJ
C₂H₅OH + 3O₂ 2CO₂ + 3H₂O = – 1368.0 kJ

The standard enthalpy of formation of benzene is

A
– 321.84 kJ

B
+ 321.84 kJ

C
– 3218.4 kJ

D
+ 3218.4 kJ

Solution

of C₆H₆(benzene) = – 3218 kJ
Question 14
1 / -0

There are two crystalline forms of lead oxide (PbO) which are known as red and yellow oxides. The for these two oxides is – 219.0 kJ and – 217.3 kJ, respectively.

What is the value of when yellow form of oxide undergoes solid-solid phase transition to form red oxide?

A
1.7 kJ

B
– 1.7 kJ

C
170 kJ

D
– 17.0 kJ

Solution

PbO (Yellow) PbO (Red)
= 1.7 kJ
Question 15
1 / -0

There are two crystalline forms of lead oxide (PbO), which are named as Red and Yellow oxides.
The for these two oxides are – 219.0 kJ and – 217.3 kJ, respectively.

The , when 2 moles of red oxide undergo solid-solid phase transition to form 2 moles of yellow oxide, is

A
– 1.7 kJ

B
+ 2.4 kJ

C
– 3.4 kJ

D
+ 1.7 kJ

Solution

Using the data, the for
2 PbO (Red) 2 PbO (Yellow) = – 3.4 kJ