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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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When 1 mole of an ideal gas to 20 atm pressure and 15 L volume expands such that the final pressure becomes 10 atm and the final volume become 60 L. Calculate entropy change for the reaction (C_p.m = 30.96)

$80.2 J k^{- 1} m o l^{- 1}$

$62.42 J k^{- 1} m o l^{- 1}$

$120 \times 10^{2} J k^{- 1} m o l^{- 1}$

$27.22 J k^{- 1} m o l^{- 1}$

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Explanation

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}; \frac{T_{2}}{T_{1}} = \frac{6}{3} ∆ S = 2.303 \times n \{C_{P} \log_{10} \frac{T_{2}}{T_{1}} + R \log_{10} \frac{P_{1}}{P_{2}}\} ∆ S = 2.303 \times 1 \{30.96 \log_{10} \frac{6}{3} + R \log_{10} \frac{20}{10}\} ∆ S = 27.22 J k^{- 1} m o l^{- 1}$

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If a process is both endothermic and spontaneous, then :

$∆ S > 0$

$∆ S < 0$

$∆ H < 0$

$∆ G > 0$

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Explanation

As $∆ G = ∆ H - T ∆ S$

For spontaneous process, $∆ G = n e g a t i v e$

For endothermic process, $∆ H = p o s i t i v e$

Therefore $∆ S > 0$

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The bond energies of C=C and C-C at 298 K are 590 and 331 kJ mol^-1 respectively. The enthalpy of polymerization per mole of ethylene is

-70 kJ

-72 kJ

72 kJ

-68 kJ

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Explanation

Polymerization reaction

$n C H_{2} = C H_{2} \to {(- C H_{2} - C H_{2} -)}_{n}$

one mole of C=C bond is broken and two moles of C-C bonds are formed per mole of ethylene.

$∆ H = 590 - 2 \times 331 = 590 - 662$

= -72 kJ per mole of ethylene.

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Which of the following statements is correct with regard to $∆$ G of a cell reaction and EMF of the cell (E) in which the reaction occurs ?

Both $∆$ G and E are extensive properties

$∆$ G is an intensive property but E is an extensive property

$∆$ G is an extensive property and E is an intensive property

Both $∆$ G and E are intensive properties.

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Explanation

(C) $∆$ G depends upon the amount of the material produced (i.e., extensive) while E is an intensive property as it is independent of the size of the cell in which the reaction is occurring.

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Temperature of 1 mol of a gas is increased by $1^{\circ}$ at constant pressure. Work done is-

R/2

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Explanation

Temperature and volume are related for adiabatic process is as

$W = P (∆ V) P V = R T P (V + ∆ V) = R (T + 1) ∴ P ∆ V = R (T + 1) - R T = R$

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