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Chemistry MCQs for NEET — Practice Questions with Answers

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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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From the following data of $∆ H$ , of the following reactions,

$C (s) + \frac{1}{2} O_{2} \to C O (g); ∆ H = - 110 k J C (s) + H_{2} O \to C O (g) + H_{2} (g); ∆ H = 132 k J$

What is the mole composition of the mixture of steam and oxygen on being passed over coke at 1273 K, keeping temperature constant.

0.5 : 1

0.6 : 1

0.8 : 1

1 : 1

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Explanation

The first reaction is exothermic and the second one is endothermic. If a mixture of steam and O₂ is passed over coke and temperature is constant, the conversion of each to CO should not show any heat change, i.e., total heat evolved in I = total heat absorbed in II.

$∴ n_{1} \times 2 \times 100 = n_{2} 132 ∴ \frac{n_{1}}{n_{2}} = \frac{132}{220} = \frac{0.6}{1}$

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State which of the following statements is true ?

First law of thermodynamics is not adequate in predicting the direction of the process.

In an exothermic reaction, the total enthalpy of the products is greater than that of reactants.

The standard enthalpy of diamond is zero at 298 K and 1 atm pressure.

It is possible to calculate the value of $∆ H$ for the reaction $H_{2} (g) + B r_{2} (l) \to 2 H B r (g)$ from the bond enthalpy data.

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Explanation

The statements given under (B), (C) are not correct because statement (B) refers to an endothermic reaction and the standard state for carbon is graphite under (C). For the calculation of $∆ H$ of the reaction under (D), additional data in the heat of vapourisation of Br₂(I) is necessary.

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The intermediate SiH₂ is formed in the thermal decomposition of silicon hydrides. Calculate $∆ H_{f}^{\circ}$ of SiH₂ given the following reactions

$S i_{2} H_{6} (g) + H_{2} (g) \to 2 S i H_{4} (g); ∆ H^{\circ} = - 11.7 k J / m o l S i H_{4} (g) \to S i H_{2} (g) + H_{2} (g); ∆ H^{\circ} = + 239.7 k J / m o l ∆ H_{f}, S i_{2} H_{6} (g) = + 80.3 k J m o l^{- 1}$

353 kJ/mol

321 kJ/mol

198 kJ/mol

274 kJ/mol

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Explanation

$S i_{2} H_{6} (g) + H_{2} (g) ⇌ 2 S i H_{4} (g) ∆ H = - 11.7 k J / m o l + 80.3 k J / m o l 0 x 2 x - 80.3 = - 11.7, ∴ 2 x = 80.3 - 11.7 = 68.6 k J / m o l ∴ x = 34.3 k J / m o l S i H_{4} (g) \to S i H_{2} (g) + H_{2} (g); ∆ H = + 239.7 k J / m o l + 34.3 y 0 y - 34.3 = 239.7 ∴ y = (239.7 + 34.3) k J m o l^{- 1} = 274 k J / m o l$

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