Practice Start Test

Thermodynamics

NEET Chemistry · 93 questions · Page 5 of 10 · Click an option or "Show Solution" to reveal answer

Q41
The values of Δ\Delta H and Δ\Delta S for the reaction, C (graphite) + CO 2 (g) → \to 2CO (g) are 170 kJ and 170 J K −-1 , respectively. This reaction will be spontaneous at
A 910 K
B 1110 K
C 510 K
D 710 K
Correct Answer
Option B
Solution

We know,

Δ\Delta

G =

Δ\Delta

H – T

Δ\Delta

S For reaction to be spontaneous,

Δ\Delta

G < 0 ⇒\Rightarrow

Δ\Delta

H – T

Δ\Delta

S < 0 ⇒\Rightarrow 170 ×\times 10 3 - T(170) < 0 ⇒\Rightarrow T > 1000 K Among the given temperatures, only 1110 K is greater than 1000 K thus, at this temperate the reaction will be spontaneous.

Q42
For the gas phase reaction, PCl 5(g) ⇌\rightleftharpoons PCl 3(g) + Cl 2(g) which of the following conditions are correct ?
A Δ\Delta H < 0 and Δ\Delta S < 0
B Δ\Delta H > 0 and Δ\Delta S < 0
C Δ\Delta H = 0 and Δ\Delta S < 0
D Δ\Delta H > 0 and Δ\Delta S > 0
Correct Answer
Option D
Solution
Δ\Delta

H =

Δ\Delta

E +

Δ\Delta

n g RT

Δ\Delta

n g = (1 + 1) – (1) = 1 ⇒\Rightarrow

Δ\Delta

H =

Δ\Delta

E + RT Thus,

Δ\Delta

H is a positive quantity i.e., ∆H > 0.

Now one mole of gaseous reactant dissociate into two moles of gaseous products thus, entropy increases i.e.,

Δ\Delta

S > 0.

Q43
Bond dissociation enthalpy of H 2 , Cl 2 and HCl are 434, 242 and 431 kJ mol −-1 respectively. Enthalpy of formation of HCl is
A −- 93 kJ mol −-1
B 245 kJ mol −-1
C 93 kJ mol −-1
D −- 245 kJ mol −-1
Correct Answer
Option A
Solution
12{1 \over 2}

H 2 +

12{1 \over 2}

Cl 2 →\to HCl,

Δ\Delta

H f = ?

Δ\Delta

H reaction = [

12{1 \over 2}

(B.E) H 2 +

12{1 \over 2}

(B.E) Cl 2 ] - (B.E) HCl = [217 + 121] – 431 = 338 – 431 = – 93 kJ mol –1

Q44
Which of the following are not state functions ? (I) q + w (II) q (III) w (IV) H −- TS
A (I), (II) and (III)
B (II) and (III)
C (I) and (IV)
D (II), (III) and (IV)
Correct Answer
Option B
Solution

Enthalpy (H = q + W) and Gibbs energy, (G = H –TS) are state functions which depend only on the initial and final states of system.

While, heat (q) and work done (W) are the path function which depends on the path followed in bringing the change between two states of the system.

Q45
Given that bond energies of H −- H and Cl −- Cl are 430 kJ mol −-1 and 240 kJ mol −-1 respectively and Δ\Delta H f for HCl is −- 90 kJ mol −-1 , bond enthalpy of HCl is
A 380 kJ mol −-1
B 425 kJ mol −-1
C 245 kJ mol −-1
D 290 kJ mol −-1
Correct Answer
Option B
Solution
12{1 \over 2}

H 2 +

12{1 \over 2}

Cl 2 →\to HCl

Δ\Delta

H f = –90 kJ mol –1 ⇒\Rightarrow

Δ\Delta

H f = [

12{1 \over 2}

(B.E) H 2 +

12{1 \over 2}

(B.E) Cl 2 ] - (B.E) HCl ⇒\Rightarrow -90 = [

12{1 \over 2}

(430) H 2 +

12{1 \over 2}

(240) Cl 2 ] - (B.E) HCl ⇒\Rightarrow -90 = [215 + 120] - (B.E) HCl ⇒\Rightarrow (B.E) HCl = 425 kJ mol –1

Q46
Consider the following reactions : (i) H + (aq) + OH −- (aq) = H 2 O (l) , Δ\Delta H = −- X 1 kJ mol −-1 (ii) H 2(g) + 1/2O 2(g) = H 2 O (l) , Δ\Delta H = −- X 2 kJ mol −-1 (iii) CO 2(g) + H 2(g) = CO (g) + H 2 O (l) , Δ\Delta H = −- X 3 kJ mol −-1 (iv) C 2 H 2(g) + 5/2O 2(g) = 2CO 2(g) + H 2 O (l) , Δ\Delta H = +X 4 kJ mol −-1 Enthalpy of formation of H 2 O (l) is
A +X 3 kJ mol −-1
B −-X 4 kJ mol −-1
C +X 1 kJ mol −-1
D −-X 2 kJ mol −-1
Correct Answer
Option D
Solution

Chemical equation for the formation of H 2 O(l) is H 2(g) + 1/2O 2(g) = H 2 O (l) Because enthalpy of formation of a compound is the heat absorbed or released when one mole of this substance is formed from its constituent elements.

Thus, enthalpy of formation of H 2 O is –X 2 kJ mol –1 where negative sign shows that the reaction is exothermic.

Equation (i) represents nutralisation reaction.

Equation (iii) represents hydrogenation reaction.

Equation (iv) represents combustion reaction.

Q47
Assume each reaction is carried out in an open container. For which reaction will Δ\Delta H = Δ\Delta E ?
A 2CO (g) + O 2(g) → \to 2CO 2(g)
B H 2(g) + Br 2(g) → \to 2HBr (g)
C C (s) + 2H 2 O (g) → \to 2H 2(g) + CO 2(g)
D PCl 5(g) → \to PCl 3(g) + Cl 2(g)
Correct Answer
Option B
Solution

We know that

Δ\Delta

H =

Δ\Delta

E +

Δ\Delta

n g RT In the reaction, H 2 (g) + Br 2 (g) →\to 2HBr(g)

Δ\Delta

n g = 2 - (1 + 1) = 0 So,

Δ\Delta

H =

Δ\Delta

E for this reaction.

Q48
Identify the correct statement for change of Gibb's energy for a system (Δ\Delta G system ) at constant temperature and pressure.
A If Δ\Delta G system < 0, the process is not spontaneous.
B If Δ\Delta G system > 0, the process is spontaneous.
C If Δ\Delta G system = 0, the system has attained equilibrium.
D If Δ\Delta G system = 0, the system is till moving in a particular direction.
Correct Answer
Option C
Solution
Δ\Delta

G system < 0, process is spontaneous.

Δ\Delta

G system = 0, process is in equilibrium.

Δ\Delta

G system > 0, process is not spontaneous.

Q49
The enthalpy of hydrogenation of cyclohexene is is −- 119.5 kJ mol −-1 . If resonance energy of benzene is −- 150.4 kJ mol −-1 , its enthalpy of hydrogenation would be
A −- 358.5 kJ mol −-1
B −- 508.9 kJ mol −-1
C −- 208.1 kJ mol −-1
D −- 269.9 kJ mol −-1
Correct Answer
Option C
Solution

The resonance energy provides extra stability to the benzene molecule so it has to over come for hydrogenation to take place.

So

Δ\Delta

H = – 358.5 – (–150.4) = –208.1 kJ

Q50
The enthalpy and entropy change for the reaction: Br 2(l) + Cl 2(g) → \to 2BrCl (g) are 30 kJ mol −-1 and 105 J K −-1 mol −-1 respectively. The temperature at which the reaction will be in equilibrium is
A 300 K
B 285.7 K
C 273 K
D 450 K
Correct Answer
Option B
Solution
Δ\Delta

G =

Δ\Delta

H – T

Δ\Delta

S Now, at equilibrium

Δ\Delta

G = 0 0 =

Δ\Delta

H – T

Δ\Delta

S ⇒\Rightarrow 0 = 30000 –T (105) ⇒\Rightarrow T =

30000105{{30000} \over {105}}

= 285.7 K

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