Thermodynamics — Page 7 | NEET Chemistry

Q61

Formation of solution from two components can be considered as (i) Pure solvent

\to

separated solvent molecules,

\Delta

H 1 (ii) Pure solute

\to

separated solute molecules,

\Delta

H 2 (iii) Separated solvent and solute molecules

\to

solution,

\Delta

H 3 Solution so formed will be ideal if

A

\Delta

H soln =

\Delta

H 1 +

\Delta

H 2 +

\Delta

H 3

B

\Delta

H soln =

\Delta

H 1 +

\Delta

H 2

-

\Delta

H 3

C

\Delta

H soln =

\Delta

H 1

-

\Delta

H 2

-

\Delta

H 3

D

\Delta

H soln =

\Delta

H 3

-

\Delta

H 1

-

\Delta

H 2

Correct Answer

Option A

Solution

For ideal solution,

\Delta

H soln =

\Delta

H 1 +

\Delta

H 2 +

\Delta

H 3

Q62

The densities of graphite and diamond at 298 K are 2.25 and 3.31 g cm

-

3 , respectively. If the standard free energy difference

\left( {\Delta {G^o}} \right)

is equal to 1895 J mol

-

1 , the pressure at which graphite will be transformed into diamond at 298 K is

A 11.14

\times

10 8 Pa

B 11.14

\times

10 7 Pa

C 11.14

\times

10 6 Pa

D 11.14

\times

10 5 Pa

Correct Answer

Option A

Solution

C (graphite) $\to$ C (diamond) Volume of graphite =

{{12} \over {2.25}} = 5.33

cm 3 mol –1 Volume of diamond =

{{12} \over {3.31}} = 3.63

cm 3 mol –1

\Delta

V = V graphite – V diamond = 1.70 cm 3 mol –1 = 1.70 × 10 –3 L mol –1 So,

\Delta

G o = P

\Delta

V $\Rightarrow$ 1895 J mol –1 = P(1.70 × 10 –3 L mol –1 ) $\Rightarrow$ 1114.70 × 10 3 J/L = P $\Rightarrow$

{{1114.7 \times {{10}^3}} \over {101.33}}

= P $\Rightarrow$ P = 11000.69 atm × 1.013 × 10 5 Pa = 11143.69 × 10 5 Pa = 11.14 × 10 8 Pa

Q63

For the reaction, C 3 H 8(g) + 5O 2(g)

\to

3CO 2 (g) + 4H 2 O (l) at constant temperature,

\Delta

H

-

\Delta

E is

A + RT

B

-

3RT

C + 3RT

D

-

RT

Correct Answer

Option B

Solution

\Delta

H =

\Delta

E +

\Delta

n g RT

\Delta

n g = 3 – (1 + 5) = – 3 $\Rightarrow$

\Delta

H =

\Delta

E + (– 3RT) $\Rightarrow$

\Delta

H –

\Delta

E = – 3RT

Q64

Heat of combustion

\Delta

H o for C (s) , H 2(g) and CH 4(g) are

-

94,

-

68 and

-

213 kcal/mol, then

\Delta

H o for C (s) + 2H 2(g)

\to

CH 4(g) is

A

-

17 kcal

B

-

111 kcal

C

-

170 kcal

D

-

85 kcal

Correct Answer

Option A

Solution

C (s) + 2H 2(g) $\to$ CH 4(g) ,

\Delta

H o = ? C (s) + O 2 (g) → CO 2(g) ,

\Delta

H = – 94 kcal/mol ....(1) H 2(g) +

{1 \over 2}

O 2 (g) $\to$ H 2 O (g) ,

\Delta

H = – 68 kcal/mol ....(2) CH 4 (g) + 2O 2 (g) $\to$ CO 2(g) + H 2 O (l) ,

\Delta

H = – 213 kcal/mol ....(3) Performing (1) + 2 × (2) – (3)

\Delta

H o = [– 94 – 2(68)] – (–213) kcal/mol = [– 94 – 136] + 213 k cal/mol = – 230 + 213 = – 17 kcal/mol

Q65

In a closed insulated container a liquid is stirred with a paddle to increase the temperature which of the following is true ?

A

\Delta

E = W

\ne

0, q = 0

B

\Delta

E = w = q

\ne

0

C

\Delta

E = 0, W = q

\ne

0

D W = 0,

\Delta

E = q

\ne

0.

Correct Answer

Option A

Solution

When in a closed insulated container, a liquid is stirred, it increases the temperature of liquid thus it started behaving as adiabatic process resulting no change in heat and only temperature increases.

Therefore, q = 0 for this process.

Also,

\Delta

E = q + W

\Delta

E = W (As, q = 0) As, there is rise in temperature thus, there must be change in internal energy, i.e.,

\Delta

E = W ≠ 0.

Q66

Unit of entropy is

A J K

-

1 mol

-

1

B J mol

-

1

C J

-

1 K

-

1 mol

-

1

D J K mol

-

1

Correct Answer

Option A

Solution

Entropy is the change in heat per degree. Thus, its unit is as follows. Entropy =

{{Jmo{l^{ - 1}}} \over K}

= J mol -1 K -1

Q67

2 mole of ideal gas at 27 o C temperature is expanded reversibly from 2 lit. to 20 lit. Find entropy change. (R = 2 cal/mol K)

A 92.1

B 0

C 4

D 9.2

Correct Answer

Option D

Solution

For isothermal reversible expansion w = q = nRT $\times$ 2.303

\log {{{v_2}} \over {{v_1}}}

= 2RT $\times$ 2.303

\log {{{20}} \over {{2}}}

= 2 × 2 × T × 2.303 × 1 = 9.2 T Entropy change,

\Delta

S =

{q \over T}

=

{{9.2T} \over T}

= 9.2 cal/mol K

Q68

Which reaction is not feasible?

A 2KI + Br 2

\to

2KBr + I 2

B 2KBr + I 2

\to

2KI + Br 2

C 2KBr + Cl 2

\to

2KCl + Br 2

D 2H 2 O + 2F 2

\to

4HF + O 2

Correct Answer

Option B

Solution

2KBr + I 2 $\to$ 2KI + Br 2 reaction is not possible because Br – ion is not oxidised in Br 2 with I 2 due to higher electrode (oxidation) potential of I 2 than bromine.

Q69

Enthalpy of CH 4 +

{1 \over 2}

O 2

\to

CH 3 OH is negative. If enthalpy of combustion of CH 4 and CH 3 OH are x and y respectively. Then which relation is correct?

A x > y

B x < y

C x = y

D x 3 y

Correct Answer

Option B

Solution

CH 4 +

{1 \over 2}

O 2 $\to$ CH 3 OH Given that for this reaction,

\Delta

H r = -ve CH 4 + 2O 2 → CO 2 + 2H 2 O,

\Delta

H = x ....(1) CH 3 OH +

{3 \over 2}

O 2 → CO 2 + 2H 2 O,

\Delta

H = y ....(2) Eqn. (1) – Eqn. (2) CH 4 +

{1 \over 2}

O 2 $\to$ CH 3 OH

\Delta

H r = x - y = -ve $\therefore$ x < y

Q70

Change in enthalpy for reaction, 2H 2 O 2(l)

\to

2H 2 O (l) + O 2(g) if heat of formation of H 2 O 2(l) and H 2 O (l) are

-

188 and - 286 kJ/mol respectively, is

A

-

196 kJ/mol

B + 196 kJ/mol

C +948 kJ/mol

D

-

948 kJ/mol

Correct Answer

Option A

Solution

2H 2 O 2(l) $\to$ 2H 2 O (l) + O 2(g) ,

\Delta

H r = ? H 2(g) + O 2(g) → H 2 O 2(l) ,

\Delta

H = – 188 kJ mol ...(1) H 2(g) + O 2(g) → H 2 O (l) ,

\Delta

H = –286 kJ/mol ....(2) (1) - (2) H 2 O 2(l) → H 2 O (l) +

{1 \over 2}

O 2(g)

\Delta

H = –286 – (– 188) = –98 kJ mol –1 Multiplying this equation by 2 we get the required equation 2H 2 O 2(l) → 2H 2 O (l) + O 2(g)

\Delta

H = 2 (– 98) kJ/mol = – 196 kJ/mol