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Chemical Thermodynamics MCQs for NEET — Chemistry Questions with Answers

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At 27°C, one mole of an ideal gas is compressed isothermally and reversibly from a pressure of 2 atm to 10 atm. The values of ΔE and q are (R = 2) [BHU 2001]

0, – 965.84 cal

– 965.84 cal, + 965.84 cal

+ 865.58 cal, – 865.58 cal

– 865.58 cal, – 865.58 cal

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Explanation

$W = 2 .303 n R T \log \frac{P_{2}}{P_{1}}$

$= 2 .303 \times 1 \times 2 \times 300 \log \frac{10}{2} = 965 .84$

at constant temperature, ΔE = 0.

ΔE = q + w; q = –w = –965.84 cal.

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ΔE° of combustion of isobutylene is –X kJ mol^–1. The value of ΔH° is [DCE 2004]

= ΔE°

> ΔE°

= 0

< ΔE°

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Explanation

${(C H_{3})}_{2} C = C H_{2} (g) + 6 O_{2} (g) \to 4 C O_{2} (g) + 4 H_{2} O (l)$

$Δ n g = 4 - 6 = - 2$ (i.e., negative)

we know that $Δ H = Δ E + Δ n g R T$

$= Δ E - (Δ n g) R T$ (âˆµ Δng = –ve)

∴ ΔH < ΔE

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An ideal gas expands in volume from 1 × 10^–3 m³ to 1 × 10^–2 m³ at 300 K against a constant pressure of 1 × 10⁵ Nm^–2. The work done is [AIEEE 2004]

270 kJ

–900 kJ

–900 J

900 kJ

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Explanation

$W = - P Δ V = - 1 \times 10^{5} (1 \times 10^{- 2} - 1 \times 10^{- 3})$

$= - 1 \times 10^{5} \times 9 \times 10^{- 3} = - 900 J$

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The spontaneous flow of heat is always

From low to high pressure

From high to high pressure

Unidirectional from lower temperature to higher temperature

Unidirectional from the higher to lower temperature

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Explanation

Heat is always flow from the higher to lower temperature.

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Mixing of non-reacting gases is generally accompanied by

Decrease in entropy

Increase in entropy

Change in enthalpy

Change in free energy

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Explanation

Mixing of non-reacting gases increases randomness and so increase entropy.

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An irreversible process occuring isothermally in an isolated system leads to

Zero entropy

An increase in the total entropy of the system

A decrease in the total entropy of the system

None of these

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Explanation

Entropy of the system increases as the process occur irreversibly and isothermally in an isolated system.

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The entropy values (in JK^-1 mol^-1) of H_2(g) = 130.6, Cl_2(g) = 223.0 and HCl_(g) = 186.7 at 298 K and 1 atm pressure. Then entropy change for the reaction $H_{2 (g)} + {Cl}_{2 (g)} \to 2 {HCl}_{(g)}$ is [BHU 2005]

+ 540.3

+ 727.3

– 166.9

+ 19.8

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Explanation

$Δ S^{o} = 2 S_{H C l}^{o} - (S_{H_{2}}^{o} + S_{C l_{2}}^{o})$

$= 2 \times 186 .7 - (130 .6 + 223 .0) = 19 .8 J K^{- 1} m o l^{- 1}$

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The ΔS for the vaporisation of 1 mol of water is 88.3 J/mole K. The value of ΔS for the condensation of 1 mol of vapour will be

88.3 J/mol K

(88.3)² J/mol K

– 88.3 J/mol K

$\frac{1}{88 .3}$ J/mol K

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Explanation

For reverse reaction sign will be change.

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The occurrence of a reaction is impossible if [AIIMS 1982, 91; MP PET 1997; MP PMT 1999; DPMT 2002]

ΔH is +ve ; ΔS is also + ve but ΔH < TΔS

ΔH is – ve ; ΔS is also – ve but ΔH > TΔS

ΔH is – ve ; ΔS is + ve

ΔH is + ve ; ΔS is – ve

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Explanation

+ve ΔH and –ve ΔS both oppose the reaction.

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When enthalpy and entropy change for a chemical reaction are –2.5 × 10³ cal and 7.4 cal deg^–1 respectively. Predict the reaction at 298 K is [AFMC 1998; MH CET 1999; CBSE PMT 2000]

Spontaneous

Reversible

Irreversible

Non-spontaneous

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Explanation

When ΔH = –ve, ΔS = +ve and ΔG = –ve than reaction is spontaneous.

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