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Chemistry MCQs for NEET — Practice Questions with Answers

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Which of the following does not exhibit zero entropy at absolute zero

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Explanation

Glass does not exhibit zero entropy at absolute zero. According to the third law of thermodynamics, crystalline substances have zero entropy at absolute zero. However, glass is an amorphous solid and does not have a perfectly ordered structure, thus retaining some entropy even at absolute zero.

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The favourable conditions for a spontaneous reaction are

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On passing $CO_2$ gas in water, its entropy

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When does the reaction occur spontaneously on the basis of the relation $ OG ^\circ = –RT/nK $ ?

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Explanation

The relation $ΔG^ ext{°} = -RT ext{ln} K$ indicates that for a reaction to be spontaneous, ΔG must be negative. This happens when the equilibrium constant K is greater than 1 because the natural logarithm of a number greater than 1 is positive, making $ΔG^ ext{°}$ negative.

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In thermodynamics, a process is called reversible when,

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Explanation

A process in thermodynamics is called reversible when the surroundings are always in equilibrium with the system. This means that any infinitesimal change in the system can be reversed without leaving any net change in either the system or the surroundings. It ensures that the process can proceed in both forward and backward directions without any loss of energy.

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Under certain conditions, the value of OG for a hypothetical reaction, $ X + Y \rightarrow Z $ is greater than zero, then –

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Explanation

If the Gibbs free energy change (ΔG) for a reaction is greater than zero, the reaction is non-spontaneous under standard conditions. This means that the reaction does not have a tendency to proceed towards the product Z. Instead, the reactants X and Y would predominate in the final mixture because the reaction favors the formation of reactants rather than products.

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For which of the following processes will energy be absorbed –

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Explanation

Separating an electron from a neutral atom is an endothermic process, meaning energy is absorbed. This process requires input energy to overcome the attraction between the electron and the nucleus, thus ionizing the atom. The other processes listed involve particles that do not require such energy input under standard conditions.

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For the combustion of 1 mole of liquid benzene at $ 25 ^\circ C $ , the heat of reaction at constant pressure is given by, $ C_6H_6(l) + 7 O_2 (g) \rightarrow 6CO_2 (g) + 3H_2O (l); OH = –780980 cal $ . What would be the heat of reaction at constant volume?

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Explanation

$ OH = OE + On_g RT $ $ Here, On_g = 6 – 7.5 = – 1.5$ . $ Thus, OE = OH + On_g RT = – 780980 – (–1.5 ) \times 2 \times 298 = – 780090 calories $

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Calculate heat of the following reaction at constant pressure, $ F_2O(g) + H_2O(g) \rightarrow O_2 (g) + 2HF(g) $ The heats of formation of $F_2O (g), H_2O(g) and HF (g) $ are 5.5 kcal, –57.8kcal and 64.2 kcal respectively.

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Explanation

$ F_2O (g) + H_2O (g) \rightarrow O_2 (g) + 2HF(g); OH = 76.1 kcal. $

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Calculate OH at 358 K for the reaction $ Fe_2O_3(s) + 3H_2(g) \rightarrow 2Fe (s) + 3H_2O(l) $ Given that, $ OH_{298} = – 33.29 kJ mole–1 and Cp for Fe_2O_3 (s), Fe (s), H_2O (l) and H_2 (g)$ are 103.8, 25.1, 75.3 and 28.8 J/K mole.

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Explanation

$ \triangle C_p = 2 \times 25.1 + 3 \times 75.3 – [103.8 + 3 \times 28.8] = 85.9 J/K mole. $
We have, $ { \triangle H_2 - \triangle H_1 \over T_2 - T_1 } = \triangle C_p $ $ { \triangle H_{358} - (-33290) \over 358 -298 } = 85. 9 $ $ \triangle H _ { 358 } = -28136 J/mole = -28.136 kJ / mole $

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