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During the isothermal mixing of ideal gases at pressure, p, the entropy change per mole for the mixing process is-R $\sum x_{i} \ln x_{i}$ where x₁, x₂,....,x_i are the mole fractions of the components, 1, 2,....,i of the mixture. Assuming ideal gas behavior, calculate $∆ S$ for the mixing of 0.8 mole of N₂ and 0.2 mole of O₂.(at $25^{\circ} C$ and 0.9 atm) [1 eu = cal/deg]

0.9943 eu

0.7533 eu

0.6798 eu

0.7112 eu

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Explanation

Mole fractions are 0.8 and 0.2

$∴$ entropy of mixing per mole

$= - 1.987 [0.8 \ln (0.8) + 0.2 \ln (0.2)] = - 1.987 [- 0.1785 - 0.3219] = 0.9943 e u$

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For the reaction : $X_{2} O_{4} (l) \to 2 X O_{2} (g) ∆ E = 2.1 k c a l . ∆ S = 20 c a l / K a t 300 K . H e n c e ∆ G i s :$

2.7 kcal

-2.7 kcal

9.3 kcal

-9.3 kcal

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Explanation

$∆ H = ∆ E + ∆ n_{g} R T = 2.1 + 2 \times 0.002 \times 300 = 3.03 k c a l ∆ G = ∆ H - T ∆ S = 3.3 - 300 \times (0.02) = - 2.7 k c a l$

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Heat of hydrogenation of ethene is x₁ and that of benzene is x₂. Hence resonance energy is-

$x_{1} - x_{2}$

$x_{1} + x_{2}$

$3 x_{1} - x_{2}$

$x_{1} - 3 x_{2}$

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Explanation

$C_{6} H_{6} + 3 H_{2} = C_{6} H_{12} ∆ H = x_{2} 3 C_{2} H_{4} + 3 H_{2} = 3 C_{2} H_{6} ∆ H = x_{1}$

$∴$ Resonance energy of benzene (contains three double bond) = 3x₁-x₂

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$C_{2} H_{6} (g) + 3.5 O_{2} (g) \to 2 C O_{2} (g) + 3 H_{2} O (g) ∆ S_{v a p} (H_{2} O, l) = x_{1} c a l K^{- 1} (b . p . + T_{1}) ∆ H_{f} (H_{2} O, l) = x_{2}; ∆ H_{f} (C O_{2}) = x_{3}, ∆ H_{f} (C_{2} H_{6}) = x_{4} H e n c e ∆ H f o r t h e r e a c t i o n i s -$

$2 x_{3} + 3 x_{2} - x_{4}$

$2 x_{3} + 3 x_{2} - x_{4} + 3 x_{1} T_{1}$

$2 x_{3} + 3 x_{2} - x_{4} - 3 x_{1} T_{1}$

$x_{1} T_{1} + x_{2} + x_{3} - x_{4}$

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Explanation

$∆ H = 3 \times ∆ H_{f} (H_{2} O, g) + 2 H_{f} (C O_{2}, g) - ∆ H_{f} (C_{2} H_{6}, g) H_{2} O (l) \to H_{2} O (g) ∆ H_{f} ∆ H_{f} = ∆ H_{f} (H_{2} O, g) - ∆ H_{f} (H_{2} O, l) ∆ H_{f} = (H_{2} O, g) = T_{1} x_{1} + ∆ H_{f} (H_{2} O, l) = T_{1} x_{1} + x_{2} ∴ ∆ H = 3 T_{1} x_{1} + 3 x_{2} + 2 x_{3} - x_{4}$

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The bond energies of $C \equiv C$ , C-H, H-H and C=C are 198, 98, 103 and145 kcal respectively. The enthalpy change of the reaction $H C \equiv C H + H_{2} \to C_{2} H_{4}$ is

48 kcal

96 kcal

-40 kcal

-152 kcal

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Explanation

The enthalpy change for the given reaction is calculated by considering the number of the various bonds broken in the reactants and the number of new bonds formed in the products.

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$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)$

B.E. (H-H) = x₁; B.E. (O=O) = x₂ B.E. (O-H) = x₃

Latent heat of vaporization of water liquid into water vapour = x₄, then $∆ H_{f}$ (heat of formation of liquid water) is-

$x_{1} + \frac{x_{2}}{2} - x_{3} + x_{4}$

$2 x_{3} - x_{1} - \frac{x_{2}}{2} - x_{4}$

$x_{1} + \frac{x_{2}}{2} - 2 x_{3} - x_{4}$

$x_{1} + \frac{x_{2}}{2} - 2 x_{3} + x_{4}$

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Explanation

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g) ∆ H ∆ H = - 2 x b o n d f o r m a t i o n e n e r g y o f (O - H) + b o n d b r e a k i n g e n e r g y o f H_{2} + \frac{1}{2} b o n d b r e a k i n g e n e r g y o f O_{2} ∆ H = x_{1} + \frac{x_{2}}{2} - 2 x_{3} H_{2} O (g) \to H_{2} O (l) ∆ H_{1} = - x_{4} ∴ H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l) ∆ H_{2} ∆ H_{2} = x_{1} + \frac{x_{2}}{2} - 2 x_{3} - x_{4}$

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In a process the pressure of a gas is inversely proportional to the square of the volume. If temperature of the gas is increases, then work done on the gas-

is positive

is negative

is zero

may be positive

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Explanation

$P \propto \frac{1}{V^{2}} \Rightarrow P = \frac{k}{V^{2}} \Rightarrow P V^{2} = k P V . V = k \Rightarrow n R T V = k \Rightarrow T V = k_{1}$

Since temperature increases therefore volume decreases.

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The enthalpies of formation of CO₂(g) and CO(g) at 298 K are in the ratio 2.57 : 1. For the reaction,

$C O_{2} (g) + C (s) \to 2 C O (g), ∆ H = 172.5 k J, ∆ H_{f} o f C O (g) i s$

-150.6 kJ mol^-1

-109.8 kJ mol^-1

-130.2 kJ mol^-1

-141.8 kJ mol^-1

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Explanation

$\frac{∆ H_{C O_{2}}}{∆ H_{C O}} = 2.57; ∆ H_{C O_{2}} = 2.57 ∆ H_{C O} 172.5 = ∆ H_{C O} - ∆ H_{C O_{2}} = ∆ H_{C O} - 2.57 ∆ H_{C O} ∆ H_{C O} = - \frac{172.5}{1.57} = - 109.8 k J m o l^{- 1}$

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In an adiabatic expansion the product of pressure and volume-

decreases

increases

remains constant

first increase then decreases

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Explanation

In an adiabatic expansion, internal energy decreases and hence temperature decreases from equation of state of ideal gas, PV=nRT

$\Rightarrow$ The product of P and V decreases.

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The molar entropy of vapourisation of acetic acid is 144.4 cal K^-1 mol^-1 at its boiling point $118^{\circ} C$ . The latent heat of vapourisation of acetic acid is

49 cal g^-1

64 cal g^-1

94 cal g^-1

84 cal g^-1

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Explanation

$∆ s = \frac{∆ H}{T (K)}; 144.4 = \frac{∆ H}{391} ∆ H = 391 \times 144.4 = 5630 c a l . m o l^{- 1} = \frac{5630}{60} = 94 c a l . g^{- 1}$

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