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Chemical Thermodynamics MCQs for NEET — Chemistry Questions with Answers

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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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When 0.16 g of glucose was burnt in a bomb calorimeter, the temperature rose by 4 deg. Calculate the calorimeter constant (water equivalent of the calorimeter) given that $∆ H^{\circ} = - 2.8 \times 10^{6} J m o l^{- 1} .$ [molar enthalpy of combustion]. Molar mass of glucose = 180 mol^-1.

$5.73 \times 10^{2} J / d e g$

$7.53 \times 10^{2} J / d e g$

$6.22 \times 10^{2} J / d e g$

$3.57 \times 10^{2} J / d e g$

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Explanation

180 gms of glucose $\Rightarrow$ $2.8 \times 10^{6} J$ of heat evolved

$∴$ 0.16 gms would yield $\frac{2.8 \times 10^{6}}{180} \times 0.16 J$

If the calorimeter constant = W, then

$W \times 4 = \frac{2.8 \times 0.16 \times 10^{6}}{180} = \frac{2.8 \times 1.6 \times 10^{3}}{1.8} J ∴ W = \frac{2.8 \times 1.6 \times 10^{3}}{1.8 \times 4} J d e g^{- 1} = 6.22 \times 10^{2} J d e g^{- 1}$

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The C-Cl bond energy can be calculated from :

$∆ {H^{\circ}}_{f} (C C l_{4}, l) o n l y$

${∆^{\circ}}_{f} (C C l_{4}, l) a n d ∆ (C l_{2})$

$∆ {H^{\circ}}_{f} (C C l_{4}, l) ∆ (C l_{2})$

$∆ {H^{\circ}}_{f} (C C l_{4}, l) ∆ (C l_{2}), ∆ {H^{\circ}}_{f} (C, g) a n d ∆ {H^{\circ}}_{v a p} (C C l_{4})$

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Explanation

$C (s) + 2 C l_{2} (g) \to C C l_{4} (l) ∆ H_{f} (C C l_{4}, l) = ∆ H^{\circ} [C (s) \to C (g)] + 2 {(B E)}_{C l - C l} - [∆ {H^{\circ}}_{v a p} (C C l_{4}) + 4 {(B E)}_{C l - C l}]$

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Given $∆ {H^{\circ}}_{f}$ of DyCl₃ (s) = -994.30 kJ mol^-1

$\frac{1}{2} H_{2} (g) + \frac{1}{2} C l_{2} (g) \overset{+ a q}{\to} H C l (a q . 4 M); ∆ H^{\circ} = - 158.31 k J m o l^{- 1} D y C l_{3} (s) \to_{H C l}^{a q} D y C l_{3} (a q . i n 4 M H C l); ∆ H^{\circ} = - 180.06 k J m o l^{- 1} D y (s) \to_{a q . 4 M}^{+ 3 H C l} D y C l_{3} (a q . 4 M H C l) + \frac{3}{2} H_{2} (g); ∆ H^{\circ} = x, c a l c u l a t e x .$

-966.5 kJ/mol

-699.43 kJ/mol

-596.6 kJ/mol

-569.6 kJ/mol

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Explanation

$D y (s) + \frac{3}{2} C l_{2} (g) ⇌ D y C l_{3} (s) ∆ H = - 994.30 k J m o l^{- 1} D y C l_{3} (s) \overset{a q . H C l}{\to} D y C l_{3} (a q . i n 4.0 M H C l) ∆ H = - 180.06 k J m o l^{- 1} 3 H C l (a q . 4 M) \to \frac{3}{2} H_{2} (g) + \frac{3}{2} C l_{2} (g) ∆ H = 3 \times 158.31 K J m o l^{- 1} D y (s) + 3 H C l ⇌ D y C l_{3} + \frac{3}{2} H_{2} (g) ∆ H = - 699.43 k J m o l^{- 1} = x (a q . 4 M) (a q . 4 M H C l)$

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1 g H₂ gas at S.T.P is expanded so that volume is doubled. Hence work done is:

22.4 L atm

5.6 L atm

11.2 L atm

44.8 L atm

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Explanation

V₁(volume of 1 g H₂) = 11.2 L at NTP

V₂(volume of 1 g H₂) = 22.4 L

$∴ W = P ∆ V = 11.2 L a t m$

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$∆ H$ for the reaction 2C(s) + 3H₂(g) $\to$ C₂H₆(g) is -20.24 kcal/mol. To what value of the enthalpy of sublimation of C(s) does this point given that the bond energies of C-C, C-H and H-H are 63 kcal/mol, 85.6 kcal/mol and 102.6 kcal/mol.

124.3 kcal/mol

185.3 kcal/mol

158.3 kcal/mol

211.5 kcal/mol

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