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Chemical Thermodynamics MCQs for NEET — Chemistry Questions with Answers

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The dissociation energy of CH₄ and C₂H₆ are respectively 360 and 620 Kcal/mole. The bond energy of C-C is-

260 Kcal/mole

180 Kcal/mole

130 Kcal/mole

80 Kcal/mole

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Explanation

$C H_{4} \to C (s) + 2 H_{2} (g) ∆ H = 360 K c a l C_{2} H_{6} \to 2 C (s) + 3 H_{2} (g) ∆ H = 620$

The avg C-H bond energy = $\frac{360}{4}$ = 90 Kcal

$∴$ C-C bond energy = 620 - 90 $\times$ 6 = 620 - 540 = 80 Kcal

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For an endothermic reaction-

$\sum_{P r o d u c t s} H = \sum_{R e a c \tan t s} H$

$\sum_{P r o d u c t s} H < \sum_{R e a c \tan t s} H$

$\sum_{P r o d u c t s} H > \sum_{R e a c \tan t s} H$

$\sum_{P r o d u c t s} H = 0 b u t \sum_{R e a c \tan t s} H i s p o s i t i v e .$

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Explanation

For an endothermic reaction, $∆ H$ is positive

= $\sum H (p r o d u c t s) - \sum H (r e a c \tan t s) .$

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All the natural process in this universe produce

A decrease in entropy of the universe

An increase in entropy of the universe

No change in entropy

Sometimes increase or sometimes decrease in entropy

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Explanation

For any natural process there is always an increase in entropy.

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For the reaction, $2 N_{2} (g) + O_{2} (g) \to 2 N_{2} O$ , at 298K $∆ H$ is 164 KJ mol^-1. The $∆ E$ of the reaction is

166.5 kJ mol^-1

161.5 kJ mol^-1

164.0 kJ mol^-1

169 kJ mol^-1

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Explanation

$∆ H = ∆ E + ∆ n R T . ∆ n = - 1 164 \times 10^{3} = ∆ E - 1 \times 08.314 \times 298 = ∆ E - 2.478 \times 10^{3} ∆ E = 164 \times 10^{3} + 2.478 \times 10^{3} = 166.5 K J m o l^{- 1}$

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For $A \to B$ , $∆ H = 4 k c a l m o l^{- 1}, ∆ S = 10 c a l m o l^{- 1} K^{- 1} .$ Reaction is spontaneous when temperature can be :

400 K

300 K

500 K

none is correct

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Explanation

$∆ G = - v e$ for spontaneous charge

$∆ G = ∆ H - T ∆ S ∴ T ∆ S > ∆ H, T > \frac{∆ H}{∆ S} \Rightarrow T > \frac{4000}{10} \Rightarrow T > 400 K$

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44.0 kJ of heat is required to evaporate one mole of water at 298 K. If $∆ H_{f}$ of $H_{2} O (l)$ is -286 kJ mol^-1, $∆ {H^{\circ}}_{f}$ of $H_{2} O (g)$ is

-330 kJ mol^-1

+242 kJ mol^-1

-242 kJ mol^-1

-198 kJ mol^-1

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Explanation

$H_{2} O (l) \to H_{2} O (g); ∆ H = 44 K J H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); ∆ {H^{\circ}}_{f} = - 286 K J A d d i n g H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g) ∆ {H^{\circ}}_{f} = - 286 + 44 = - 242 K J m o l^{- 1}$

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The molar heat capacity, $C_{v}$ of an ideal gas whose energy is that of translational motion only is

$2.98 J d e g^{- 1} m o l^{- 1}$

$12.47 J d e g^{- 1} m o l^{- 1}$

$6.43 J d e g^{- 1} m o l^{- 1}$

$9.41 J d e g^{- 1} m o l^{- 1}$

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Explanation

$E_{t r} = \frac{3}{2} R T f o r 1 m o l e C_{v} = {(\frac{\partial E}{\partial T})}_{v} = (\frac{\frac{3}{2} R T}{\partial T}) = \frac{3}{2} R = 12.47 J d e g^{- 1} m o l^{- 1}$

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The lattice energy of NaCl is -780 kJ mol^-1. The enthalpies of hydration of Na⁺(g) and Cl^-(g) ions are -406 kJ mol^-1 and -364 kJ mol^-1. The enthalpy of solution of NaCl(s) is

738 kJ mol^-1

10 kJ mol^-1

-10 kJ mol^-1

-822 kJ mol^-1

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Explanation

$N a C l (s) \to N a^{+} (g) + C l^{- 1} (g) ∆ H = + 780 k J N a^{+} (g) + a q \to N a^{+} (a q) ∆ H = - 406 k J C l^{-} (g) + a q \to C l^{-} (a q) ∆ H = - 364 k J$

The net reaction is dissolution of NaCl(s).

$∆ H_{s o l u t i o n} = 780 - (406 + 364) = 10 k J m o l^{- 1}$

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Enthalpy of fusion of a liquid is 1.435 kcal mol^-1 and molar entropy change is 5.26 cal mol^-1K^-1. Hence melting point of liquid is :

$100^{\circ} C$

$0^{\circ} C$

373 K

$- 273^{\circ} C$

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Explanation

$∆ S = \frac{∆ H}{T} \Rightarrow T = \frac{∆ H}{∆ S} \Rightarrow \frac{1435 c a l}{5.26} = 273 K \Rightarrow 0^{\circ} C$

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Following reaction occurs at $25^{\circ} C$ :

$2 N O (g, 1 \times 10^{- 5} a t m) + C l_{2} (g, 1 \times 10^{- 2} a t m) ⇌ 2 N O C l (g, 1 \times 10^{- 2} a t m) ∆ G^{\circ}$ is-

-45.65 kJ

-28.53 kJ

-22.82 kJ

-57.06 kJ

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Explanation

$K_{e q} = \frac{P_{N O C l}^{2}}{P_{N O}^{2} \times P_{C l_{2}}} = 10^{8} ∴ ∆ G^{\circ} = - R T \ln K_{e q} = - 2.303 \times 8.314 \times 298 \times 8 = - 45.65 k J$

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