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Chemical Thermodynamics MCQs for NEET — Chemistry Questions with Answers

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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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100 ml of 0.3 M HCl solution is mixed with 100 ml of 0.35 M NaOH solution. The amount of heat liberated is

7.3 kJ

5.71 kJ

10.42 kJ

1.713 kJ

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Explanation

Amount of HCl added = $\frac{100 \times 0.3}{1000}$ = 0.03 mol

Amount of NaOH added = $\frac{100 \times 0.35}{1000}$ = 0.035 mol

Limiting reagent = 0.03 mol of HCl

$∴$ amount of heat evolved = $57.1 \times 0.03 = 1.713 k J$

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Five moles of ideal gas expand isothermally and reversibly from an initial pressure of 100 atm to a final pressure of 1 atm at $27^{\circ} C$ . The work done by the gas is

3455 cal

6909 cal

69010 cal

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Explanation

The work done in the isothermal reversible expansion of n moles of an ideal gas is given by the equation.

$W = n R T \ln \frac{V_{2}}{V_{1}} = n R T \ln \frac{P_{1}}{P_{2}}$

where V₁, V₂ are the initial and final volumes of the gas, the corresponding pressure P₁ and P₂.

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The quantity $δ_{q}$ i.e. heat absorbed an infinitesimal process is

Dependent on the path of transformation

Dependent on the thermodynamic state of the system

Independent of both

An exact differential

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Explanation

$δ_{q}$ is a path dependent property and is also an inexact differential.

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The latent heat of vapourisation of water at $25^{\circ} C$ is 10.5 kcal mol^-1 and the standard heat of formation of liquid water is -68.3 kcal. The enthalpy change of the reaction

$H_{2} (g) + (1 / 2) O_{2} (g) \to H_{2} O (g)$ is therefore,

-57.8 kcal

-78.8 kcal

78.8 kcal

-47.3 kcal

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Explanation

The result can be arrived from the equations

$H_{2} O (l) = H_{2} O (g); ∆ H 10.5 k c a l m o l^{- 1} H_{2} (g) + \frac{1}{2} O_{2} (g) = H_{2} O (l); ∆ H = 68.3 k c a l m o l^{- 1}$

by suitable algebraic manipulation.

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Which of the following thermodynamic quantities is an outcome of the second law of thermodynamics ?

Work

Enthalpy

Internal energy

Entropy

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Explanation

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The entropy concept is direct outcome of the second law of thermodynamics through the Carnot cycle approach.

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