Practice Start Test

Thermodynamics

NEET Chemistry · 93 questions · Page 6 of 10 · Click an option or "Show Solution" to reveal answer

Q51
A reaction occurs spontaneously if
A TΔ\Delta S < Δ\Delta H and both Δ\Delta H and Δ\Delta S are +ve
B TΔ\Delta S > Δ\Delta H and Δ\Delta H is +ve and Δ\Delta S are −-ve
C TΔ\Delta S > Δ\Delta H and both Δ\Delta H and Δ\Delta S are +ve
D TΔ\Delta S = Δ\Delta H and both Δ\Delta H and Δ\Delta S are +ve
Correct Answer
Option C
Solution
Δ\Delta

G =

Δ\Delta

H – T

Δ\Delta

S For spontaneous reaction,

Δ\Delta

G has to be negative. Among the given options, it is positive only when T

Δ\Delta

S >

Δ\Delta

H and both

Δ\Delta

H and

Δ\Delta

S are +ve .

Q52
Which of the following pairs of a chemical reaction is certain to result in a spontaneous reaction ?
A Exothermic and increasing disorder
B Exothermic and decreasing disorder
C Endothermic and increasing disorder
D Endothermic and decreasing disorder
Correct Answer
Option A
Solution
Δ\Delta

G =

Δ\Delta

H – T

Δ\Delta

S For a reaction to become spontaneous,

Δ\Delta

G must be negative. For case (a) exothermic and increasing disorder For exothermic reaction,

Δ\Delta

H = –ve and increasing disorder,

Δ\Delta

S = +ve

Δ\Delta

G = –ve –T(+ve) Thus,

Δ\Delta

G is negative for all temperature range For case (b) exothermic and decreasing disorder For exothermic reaction,

Δ\Delta

H = –ve and for decreasing disorder,

Δ\Delta

S = –ve ⇒\Rightarrow

Δ\Delta

G = –ve – T(–ve) Thus,

Δ\Delta

G is not negative for all temperature range. For case (c) endothermic and increasing disorder. For endothermic reaction,

Δ\Delta

H = +ve and increasing disorder,

Δ\Delta

S = +ve Thus,

Δ\Delta

G is not negative for all temperature range. For case (d) endothermic and decreasing disorder For endothermic reaction,

Δ\Delta

H = +ve and decreasing disorder,

Δ\Delta

S = –ve

Δ\Delta

G = +ve – T (–ve) Thus,

Δ\Delta

G is positive for all temperature range.

Q53
The absolute enthalpy of neutralisation of the reaction : Mg(O) (s) + 2HCl (aq) → \to MgCl 2(aq) + H 2 O (l) will be
A −-57.33 kJ mol −-1
B greater than −- 57.33 kJ mol −-1
C less than −- 57.33 kJ mol −-1
D 57.33 kJ mol −-1
Correct Answer
Option C
Solution

We know that enthalpy of neutralization of a strong acid and a strong base is –57.33 kJ mol –1 .

Here, MgO is the weak base and HCl is a strong acid thus, a small amount of energy is used in the ionization of MgO thus, the heat of neutralization decreases.

Therefore, enthalpy of neutralization is less than – 57.33 kJ mol –1

Q54
Considering entropy (S) as a thermodynamic parameter, the criterion for the spontaneity of any process is
A ΔSsystem+ΔSsurroundings>0\Delta {S_{system}} + \Delta {S_{surroundings}} > 0
B ΔSsystem−ΔSsurroundings>0\Delta {S_{system}} - \Delta {S_{surroundings}} > 0
C ΔSsystem>0   \Delta {S_{system}} > 0\,\,\,only
D ΔSsurroundings>0\Delta {S_{surroundings}} > 0 only
Correct Answer
Option A
Solution

For the reaction to be spontaneous, the total entropy of system and universe increases i.e.,

ΔSsystem+ΔSsurroundings>0\Delta {S_{system}} + \Delta {S_{surroundings}} > 0
Q55
Standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are −- 382.64 kJ mol −-1 and −- 145.6 kJ mol −-1 , respectively. Standard Gibb's energy change for the same reaction at 298 K is
A −- 221.1 kJ mol −-1
B −-339.3 kJ mol −-1
C −- 439.3 kJ mol −-1
D −- 523.2 kJ mol −-1
Correct Answer
Option B
Solution
Δ\Delta

G =

Δ\Delta

H – T

Δ\Delta

S

Δ\Delta

G = –382.64 × 103 J mol –1 – (298K) (–145.6 JK –1 mol –1 ) = –382640 + 43388.8 = – 339251.2 J mol –1 = – 339.3 kJ mol –1

Q56
The work done during the expansion of a gas from a volume of 4 dm 3 to 6 dm 3 against a constant external pressure of 3 atm is (1 L atm = 101.32 J)
A −- 6 J
B −- 608 J
C + 304 J
D −- 304 J
Correct Answer
Option B
Solution

Work done during the expansion, W = – pdV W = –3 atm (6 dm 3 – 4 dm 3 ) = – 3 atm ( 2 dm 3 ) (1 dm 3 = 1 L) = – 3 atm × 2 L = – 6 L atm As, 1 L atm = 101.32 J ∴\therefore W = – 6 × 101.32 J = – 607.92 J ≈ – 608 J

Q57
If the bond energies of H −- H, Br −- Br, and H −- Br are 433, 192 and 364 kJ mol −-1 respectively, the Δ\Delta H o for the reaction H 2 (g) + Br 2(g) → \to 2HBr (g) is
A −- 261 kJ
B +103 kJ
C +261 kJ
D −-103 kJ
Correct Answer
Option D
Solution

H 2 (g) + Br 2(g) →\to 2HBr (g) ,

Δ\Delta

H o f = ?

Δ\Delta

H o f =

Σ\Sigma

(B.E.) reactants –

Σ\Sigma

(B.E.) products = (B.E.)

H–H + (B.E.)

Br–Br –2(B.E) H-Br = [433 + 192] – 2(364) kJ mol –1 = (625 – 728) kJ mol –1 = –103 kJ mol –1

Q58
The molar heat capacity of water at constant pressure, C, is 75 J K −-1 mol −-1 . When 1.0 kJ of heat is supplied to 100 g of water which is free to expand, the increase in temperature of water is
A 1.2 K
B 2.4 K
C 4.8 K
D 6.6 K
Correct Answer
Option B
Solution

As we know, q = nC

Δ\Delta

T q = 1.0 kJ = 1000 J C = 75 JK –1 mol –1 m = 100 g ⇒ Number of moles =

10018{{100} \over {18}}

g mol –1 1000 =

10018{{100} \over {18}}

×\times 75 ×\times

Δ\Delta

T ⇒\Rightarrow

Δ\Delta

T =

10×1875{{10 \times 18} \over {75}}

K ⇒\Rightarrow

Δ\Delta

T = 2.4 K

Q59
For which one of the following equations is Δ\Delta H o react equal to Δ\Delta H o f for the product ?
A Xe (g) + 2F 2(g) → \to XeF 4(g)
B 2CO (g) + O 2(g) → \to 2CO 2(g)
C N 2(g) + O 3(g) → \to N 2 O 3(g)
D CH 4(g) + 2Cl 2(g) → \to CH 2 Cl 2(l) + 2HCl (g)
Correct Answer
Option A
Solution

Heat of formation,

Δ\Delta

H o f of a substance is the amount of heat absorbed or released when one mole of this substance is formed directly from its constituent elements.

In option (a), one mole of XeF 4 is formed from its constituent elements i.e., Xe and F 2 thus, the equation has equal value of

Δ\Delta

H o r and

Δ\Delta

H o f . In option (b), the constituent atoms should be carbon and oxygen only but the reactant used is CO thus,

Δ\Delta

H o r ≠\ne

Δ\Delta

H o f In option (c), the reactant used is O 3 which is again not in its element form thus,

Δ\Delta

H o r ≠\ne

Δ\Delta

H o f In option (d), two products are formed thus

Δ\Delta

H o r ≠\ne

Δ\Delta

H o f

Q60
What is the entropy change (in J K −-1 mol −-1 ) when one mole of ice is converted into water at 0 o C? (The enthalpy change for the conversion of ice to liquid water is 6.0 kJ mol −-1 at 0 o C).
A 20.13
B 2.013
C 2.198
D 21.98
Correct Answer
Option D
Solution

H 2 O(s) → H 2 O(l)

Δ\Delta

H = 6.0 kJ mol –1 T = 0 + 273 K = 273 K;

Δ\Delta

S = ?

Δ\Delta

S =

ΔHT{{\Delta H} \over T}

=

6000273{{6000} \over {273}}

= 21.98 JK –1 mol –1

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