Consider the following processes : H (kJ/mol) 1/2A $$ \to $$ B

H (kJ/mol) 1/2A $$ \to $$ B +150 3B $$ \to $$ 2C + D -125 E + A $$ \to $$ 2D +350 ___________________________________ B + D $$ \to $$ E + 2C; $$\Delta $$H = (300 - 125 - 350) = - 175 kJ/mol

Thermodynamics — Page 3 | NEET Chemistry

Q21

For a sample of perfect gas when its pressure is changed isothermally from p i to p f , the entropy change is given by

A

\Delta S = nR\,\ln \left( {{{{p_f}} \over {{p_i}}}} \right)

B

\Delta S = nR\,\ln \left( {{{{p_i}} \over {{p_f}}}} \right)

C

\Delta S = nRT\,\ln \left( {{{{p_f}} \over {{p_i}}}} \right)

D

\Delta S = RT\,\ln \left( {{{{p_i}} \over {{p_f}}}} \right)

Correct Answer

Option B

Solution

For an ideal gas undergoing reversible expansion, when temperature changes from T i to T f and pressure changes from P i to P f .

\Delta

S = nC p ln

{{{T_f}} \over {{T_i}}}

+ nR ln

{{{P_i}} \over {{P_f}}}

For an isothermal process, T i = T f so, ln 1 = 0 $\therefore$

\Delta

S = nR ln

{{{P_i}} \over {{P_f}}}

Q22

The heat of combination of carbon to CO 2 is

-

393.5 kJ/mol. The heat released upon formation of 35.2 g of CO 2 from carbon and oxygen gas is

A + 315 kJ

B

-

630 kJ

C

-

3.15 kJ

D

-

315 kJ

Correct Answer

Option D

Solution

To determine the heat released when 35.2 g of CO 2 is formed from carbon and oxygen gas, we will follow these steps: 1.

Calculate the number of moles of CO 2 : The molar mass of CO 2 can be calculated as follows: Carbon (C): 12.01 g/mol Oxygen (O): 16.00 g/mol So, the molar mass of CO 2 is:

12.01 + (16.00 \times 2) = 12.01 + 32.00 = 44.01 \, \text{g/mol}

The number of moles of CO 2 in 35.2 g is calculated using:

\text{Number of moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{35.2 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.8 \, \text{moles}

2.

Calculate the heat released for 0.8 moles of CO 2 : The heat of combination given is -393.5 kJ/mol.

This is the energy released when 1 mole of CO 2 is formed:

\text{Heat released} = \text{Number of moles} \times \text{Heat of combination}

\text{Heat released} = 0.8 \, \text{moles} \times -393.5 \, \text{kJ/mol} = -314.8 \, \text{kJ}

After rounding, the heat released is approximately -315 kJ. Answer: Option D $-$ 315 kJ.

Q23

Which of the following statements is correct for the spontaneous adsorption of a gas?

A

\Delta

S is negative and, therefore

\Delta

H should be highly positive.

B

\Delta

S is negative and therefore,

\Delta

H should be highly negative.

C

\Delta

S is positive and therefore,

\Delta

H should be negative .

D

\Delta

S is positive and therefore,

\Delta

H should also be highly positive.

Correct Answer

Option B

Solution

Using Gibb's-Helmholtz equation,

\Delta G = \Delta H - T\Delta S

During adsorption of a gas, entropy decreases i.e,

\Delta

S < 0 For spontaneous adsorption,

\Delta

G should be negative, which is possible when

\Delta

H is highly negative.

Q24

For the reaction,

{X_2}{O_{4\left( l \right)}}\,\, \to \,\,2X{O_{2(g)}}

\Delta

U = 2.1 kcal,

\Delta

S = 20 cal K

-

1 at 300 K Hence, G is

A 2.7 kcal

B

-

2.7 kcal

C 9.3 kcal

D

-

9.3 kcal

Correct Answer

Option B

Solution

\Delta H = \Delta U + \Delta {n_g}RT

Given

\Delta U = 2.1\,kcal,\,\Delta {n_g} = 2

R = 2 \times {10^{ - 3}}kcal,T = 300K

$\therefore$

\Delta H = 2.1 + 2 \times 2 \times {10^{ - 3}} \times 300 = 3.3\,kcal

Again,

\Delta G = \Delta H + T\Delta S

Given

\Delta

S = 20 $\times$ 10 -3 kcal K -1 On putting the values of

\Delta

H and

\Delta

S in the equation, we get

\Delta G = 3.3 - 300 \times 20 \times {10^{ - 3}}

\Rightarrow 3.3 - 6 \times {10^3} \times {10^{ - 3}} = - 2.7\,kcal

Q25

When 5 litres of a gas mixture of methane and propane is perfectly combusted at 0 o C and 1 atmosphere, 16 litres of oxygen at the same temperature and pressure is consumed, The amount of heat released from this combustion in kJ (

\Delta

H comb. (CH 4 ) = 890 kJ mol

-

1 ,

\Delta

H comb . (C 3 H 8 ) = 2220 kJ mol

-

1 ) is

A 38

B 317

C 477

D 32

Correct Answer

Option B

Solution

CH 4 + 2O 2 $\to$ CO 2 + 2H 2 O C 3 H 8 + 5O 2 $\to$ 3CO 2 + 4H2 O CH 4 + C 3 H 8 =

{5 \over {22.4}} = 0.22

moles.

{O_2} = {{16} \over {22.4}} = 0.71

moles

2x + (0.22 \times x)5 = 0.71

x = 0.13 Heat liberated = 0.13 $\times$ 890 + 0.09 $\times$ 2220 = 316 kJ

Q26

A reaction having equal energies of activation for forward and reverse reactions has

A

\Delta

H = 0

B

\Delta

H =

\Delta

G =

\Delta

S = 0

C

\Delta

S = 0

D

\Delta

G = 0

Correct Answer

Option A

Solution

For a general reaction, ΔH = Activation energy of forward reaction – Activation energy of backward reaction.

As, both the energies of activation have same value thus, ΔH = 0.

\Delta

G is not equal to zero because if it is so the reaction must be in equilibrium which is not in this case

Q27

Standard enthalpy of vaporisation

\Delta

vap H o for water at 100 o C is 40.66 kJ mol

-

1 . The internal energy of vaporisation of water at 100 o C (in kJ mol

-

1 ) is

A +37.56

B

-

43.76

C + 43.76

D + 40.66

Correct Answer

Option A

Solution

H 2 O(l)

\overset{{100^\circ \,C}}\longrightarrow

H 2 O(g) ∆H o = 40.66kJ mol –1 ∆H o = ∆u o +

\Delta

n g RT

\Delta

n g = 1, R = 8.314 × 10 –3 kJ mol –1 k –1 T = 100 + 273 = 373 K $\Rightarrow$ 40.66 = ∆u o + (1) (8.314 × 10 –3 ) × 373 $\Rightarrow$ ∆u o = 37.56 kJ mol –1

Q28

The enthalpy of fusion of water is 1.435 kcal/mol. The molar entropy change for the melting of ice at 0 o C is

A 10.52 cal/(mol K)

B 21.04 cal/(mol K)

C 5.260 cal/(mol K)

D 0.526 cal/(mol K)

Correct Answer

Option C

Solution

H 2 O(

l

) → H 2 O(s) ∆H = 1.435 Kcal/mol T = 0 + 273K = 273K

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

$\Rightarrow$

\Delta S = {{1.435} \over {273}}

= 5.26 $\times$ 10 -3 kcal/mol K $\Rightarrow$

\Delta S

= 5.260 cal/mol K

Q29

In which of the following reactions, standard reaction entropy change (

\Delta

S o ) is positive and standard Gibb's energy change (

\Delta

G o ) decreases sharply with increasing temperature ?

A C (graphite) +

{1 \over 2}

O 2(g)

\to

CO (g)

B CO (g) +

{1 \over 2}

O 2(g)

\to

CO 2(g)

C Mg (s) +

{1 \over 2}

O 2(g)

\to

MgO (g)

D

{1 \over 2}

C (graphite) +

{1 \over 2}

O 2(g)

\to

{1 \over 2}

CO 2(g)

Correct Answer

Option A

Solution

\Delta

S o positive means entropy of products is more than that of reactants. Among the given reactions only in option (a) the

\Delta

S o is positive cause products has 1 mole of gaseous products CO while reaction have half mole of gaseous O 2 and Carbon (C) in solid state.

Thus, reactions have less randomness than products.

Also, the reaction is a combustion reaction which has a negative value of

\Delta

H. According to Gibbs energy change (

\Delta

S o ).

\Delta

G o =

\Delta

H o - T

\Delta

S o $\Rightarrow$

\Delta

G o = -ve - T(+ve) Thus, on increasing temperature value of

\Delta

G o decreases sharply.

Q30

Consider the following processes : <br><br> <table class=tg> <tbody><tr> <th class=tg-wv9z></th> <th class=tg-wv9z>H (kJ/mol)</th> </tr> <tr> <td class=tg-wv9z>1/2A

\to

B</td> <td class=tg-wv9z>+150</td> </tr> <tr> <td class=tg-wv9z>3B

\to

2C + D</td> <td class=tg-wv9z>-125</td> </tr> <tr> <td class=tg-wv9z>E + A

\to

2D</td> <td class=tg-wv9z>+350</td> </tr> </tbody></table> <br>For B + D

\to

E + 2C,

\Delta

H will be

A 525 kJ/mol

B

-

175 kJ/mol

C

-

325 kJ/mol

D 325 kJ/mol

Correct Answer

Option B

Solution

\Delta

H = (300 - 125 - 350) = - 175 kJ/mol