Thermodynamics — Page 4 | NEET Chemistry

Q31

Which of the following is correct option for free expansion of an ideal gas under adiabatic condition ?

A q = 0,

\Delta

T

\ne

0, w = 0

B q

\ne

0,

\Delta

T = 0, w = 0

C q = 0,

\Delta

T = 0, w = 0

D q = 0,

\Delta

T < 0, w

\ne

0

Correct Answer

Option C

Solution

For adiabatic condition, change in heat does not take place thus, q = 0.

If q = 0 then change in temperature does not take place thus,

\Delta

T = 0. Also for a free expansion of ideal gas work done, W = 0 as there is no external pressure on it.

Q32

Enthalpy change for the reaction, 4H (g)

\to

2H 2(g) is

-

869.6 kJ The dissociation energy of H

-

H bond is

A 434.8 kJ

B

-

869.6 kJ

C + 434.8 kJ

D + 217.4 kJ

Correct Answer

Option C

Solution

4H(g) → 2H 2 (g), ∆H = – 869.6 kJ Reverse the above equation 2H 2 (g) → 4H(g), ∆H = + 869.6 kJ Divide the above equation by 2, H 2 (g) → 2H(g),

\Delta H = {{869.6} \over 2}

kJ = 434.8 kJ

Q33

If the enthalpy change for the transition of liquid water to steam is 30 kJ mol

-

1 at 27 o C, the entropy change for the process would be

A 10 J mol

-

1 K

-

1

B 1.0 J mol

-

1 K

-

1

C 0.1 J mol

-

1 K

-

1

D 100 J mol

-

1 K

-

1

Correct Answer

Option D

Solution

H 2 O(l)

\overset{{300K}}\longrightarrow

H 2 O(g) ∆H = 30 kJ mol –1

\Delta S = {{\Delta H} \over T}

=

{{30 \times {{10}^3}} \over {300}}

= 100 J mol –1 K –1

Q34

Match List I (Equations) with List II (Type of processes) and select the correct option. <br><br> <table class=tg> <tbody><tr> <th class=tg-4kyz colspan=2>List I</th> <th class=tg-bzci colspan=2>List II</th> </tr> <tr> <td class=tg-4kyz colspan=2>Equations</td> <td class=tg-bzci colspan=2>Type of processes</td> </tr> <tr> <td class=tg-13k7>A.</td> <td class=tg-4kyz>K<sub>p</sub> > Q</td> <td class=tg-60hs>(i)</td> <td class=tg-bzci>Non- spontaneous</td> </tr> <tr> <td class=tg-13k7>B.</td> <td class=tg-4kyz>

\Delta

G<sup>o</sup> < RT ln Q</td> <td class=tg-60hs>(ii)</td> <td class=tg-bzci>Equilibrium</td> </tr> <tr> <td class=tg-60hs>C.</td> <td class=tg-bzci>K<sub>p</sub> = Q</td> <td class=tg-60hs>(iii)</td> <td class=tg-bzci>Spontaneous and<br> endothermic</td> </tr> <tr> <td class=tg-60hs>D.</td> <td class=tg-bzci>T >

{{\Delta H} \over {\Delta S}}

</td> <td class=tg-60hs>(iv)</td> <td class=tg-bzci>Spontaneous</td> </tr> </tbody></table>

A A - (i), B - (ii), C - (iii), D - (iv)

B A - (iii), B - (iv), C - (ii), D - (i)

C A - (iv), B - (i), C - (ii), D - (iii)

D A - (ii), B - (i), C - (iv), D - (iii)

Correct Answer

Option C

Solution

When K p > Q, rate of forward reaction > rate of backward reaction. $\therefore$ Reaction is spontaneous. When

\Delta

G o < RT ln Q,

\Delta

G o is positive, reverse reaction is feasible, thus reaction is non spontaneous.

When K p = Q, rate of forward reaction > rate of backward reaction.

$\therefore$ Reaction is in equilibrium.

When T

\Delta

S >

\Delta

H,

\Delta

G will be negative only when

\Delta

H = +ve. $\therefore$ Reaction is spontaneous and endothermic.

Q35

The following two reactions are known Fe 2 O 3(s) + 3CO (g)

\to

2Fe (s) + 3CO 2(g) ;

\Delta

H =

-

26.8 kJ FeO (s) + CO(g)

\to

Fe (s) + CO 2(g) ;

\Delta

H =

-

16.5 kJ The value of

\Delta

H for the following reaction Fe 2 O 3(s) + CO (g)

\to

2FeO (s) + CO 2(g) is

A + 10.3 kJ

B

-

43.3 kJ

C

-

10.3 kJ

D + 6.2 kJ

Correct Answer

Option D

Solution

Given Fe 2 O 3(s) + 3CO (g) $\to$ 2Fe (s) + 3CO 2(g) ;

\Delta

H = $-$ 26.8 kJ .....(1) FeO (s) + CO(g) $\to$ Fe (s) + CO 2(g) ;

\Delta

H = $-$ 16.5 kJ .....(2) Fe 2 O 3(s) + CO (g) $\to$ 2FeO (s) + CO 2(g) ,

\Delta

H = ? ....(3) Equation (3) can be calculated as : (1) - 2(2) $\therefore$

\Delta

H = –26.8 + 33.0 = +6.2 kJ

Q36

For vaporization of water at 1 atmospheric pressure, the values of

\Delta

H and

\Delta

S are 40.63 kJ mol

-

1 and 108.8 J K

-

1 mol

-

1 , respectively. The temperature when Gibb's energy change (

\Delta

G) for this transformation will be zero, is

A 273.4 K

B 393.4 K

C 373.4 K

D 293.4 K

Correct Answer

Option C

Solution

We know, from Gibb's equation,

\Delta

G =

\Delta

H – T

\Delta

S When

\Delta

G = 0,

\Delta

H = T

\Delta

S $\therefore$

T = {{\Delta H} \over {\Delta S}}

=

{{40.63 \times {{10}^3}} \over {108.8}}

= 373.4 K

Q37

Three moles of an ideal gas expanded spontaneously into vacuum. The work done will be

A infinite

B 3 Joules

C 9 Joules

D zero

Correct Answer

Option D

Solution

Since the ideal gas expands spontaneously into vacuum,, P ext = 0. $\therefore$ Work done is also zero.

Q38

For an endothermic reaction, energy of activation is E a and enthalpy of reaction is

\Delta

H (both of these in kJ/mol). Minimum value of E a will be

A less than

\Delta

H

B equal to

\Delta

H

C more than

\Delta

H

D equal to zero

Correct Answer

Option C

Solution

Here, E a = activation energy of forward reaction E’ a = activation energy of backward reaction

\Delta

H = enthalpy of the reaction From the given diagram it is clear that E a = E’ a +

\Delta

H $\therefore$ E a >

\Delta

H

Q39

Standard entropies of X 2 , Y 2 and XY 3 are 60, 40 and 50 J K

-

1 mol

-

1 respectively. For the reaction 1/2X 2 + 3/2Y 2

\rightleftharpoons

XY 3 ,

\Delta

H =

-

30 kJ, to be at equilibrium, the temperature should be

A 750 K

B 1000 K

C 1250 K

D 500 K

Correct Answer

Option A

Solution

Given reaction is :

{1 \over 2}

X 2 +

{3 \over 2}

Y 2 ⇌ XY 3 We know,

\Delta

S o =

\sum {S_{products}^o} - \sum {S_{reac\tan ts}^o}

= 50 - (30 + 60) = -40 J K -1 mol -1 At equilibrium

\Delta

G o = 0

\Delta

H o = T

\Delta

S o $\therefore$

T = {{\Delta {H^o}} \over {\Delta {S^o}}}

=

{{ - 30 \times {{10}^3}} \over { - 40}}

= 750 K

Q40

From the following bond energies : H

-

H bond energy : 431.37 kJ mol

-

1 C

=

C bond energy : 606.10 kJ mol

-

1 C

-

C bond energy : 336.49 kJ mol

-

1 C

-

H bond energy : 410.50 kJ mol

-

1 Enthalpy for the reaction, will be

A

-

243.6 kJ mol

-

1

B

-

120.0 kJ mol

-

1

C 553.0 kJ mol

-

1

D 1523.6 kJ mol

-

1

Correct Answer

Option B

Solution

\Delta

H reaction = Σ(Bond enthalpy) reactants – Σ(Bond enthalpy) products = [B.E (C-C) + B.E (H-H) + 4 $\times$ B.E (C-H) ] - [B.E (C-C) + 6 $\times$ B.E (C-H) ] = [606.10 + 4(410.50) + 431.37] – [336.49 + 6(410.50)] = 2679.47 – 2799.49 = – 120.02 kJ mol –1