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NEET UG CHEMISTRY MCQ: THERMODYNAMICS

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Understanding +4 | -1
Consider the following reactions:
(i)   H+(aq)+OH−(aq)→H2O(l),  ΔH=−X1  kJ  mol−1\mathrm{(i)}\,\,\,{H^ + }(aq) + O{H^ - }(aq) \to {H_2}O(l),\,\,\Delta H = - {X_1}\,\,\mathrm{kJ}\,\,\mathrm{mo}{\mathrm{l}^{ - 1}}
(ii)   H2(g)+12O2(g)→H2O(l),  ΔH=−X2  kJ  mol−1\mathrm{(ii)}\,\,\,{H_2}(g) + \frac{1}{2}{O_2}(g) \to {H_2}O(l),\,\,\Delta H = - {X_2}\,\,\mathrm{kJ}\,\,\mathrm{mo}{\mathrm{l}^{ - 1}}
(iii)   CO2(g)+H2(g)→CO(g)+H2O,  ΔH=−X3  kJ  mol−1\mathrm{(iii)}\,\,\,C{O_2}(g) + {H_2}(g) \to CO(g) + {H_2}O,\,\,\Delta H = - {X_3}\,\,\mathrm{kJ}\,\,\mathrm{mo}{\mathrm{l}^{ - 1}}
(iv)   C2H2(g)+52O2(g)→2CO2(g)+H2O(l),  ΔH=+4X4  kJ  mol−1\mathrm{(iv)}\,\,\,{C_2}{H_2}(g) + \frac{5}{2}{O_2}(g) \to 2C{O_2}(g) + {H_2}O(l),\,\,\Delta H = +4 {X_4}\,\,\mathrm{kJ}\,\,\mathrm{mo}{\mathrm{l}^{ - 1}}
Enthalpy of formation of H2O(l){H_2}{O_{(l)}} is
View answer and explanation
Correct option: D
−X2  kJ  mol−1- {X_2}\,\,\mathrm{kJ}\,\,\mathrm{mo}{\mathrm{l}^{ - 1}}
Option D (−X2 kJ mol−1- {X_2}\,\mathrm{kJ}\,\mathrm{mo}{\mathrm{l}^{ - 1}}) is correct because reaction (ii) directly represents the formation of one mole of H2O(l) from its elements in their standard states.

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The internal energy of one mole of a gas is: An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If Ti{T_i} is the initial temperature and Tf{T_f} is the final temperature, which of the following statements is correct? For which of the process, ∆S is negative? The work done by a weightless piston in causing an expansion ∆V (at constant temperature), when the opposing pressure, P is variable, is given by: Assertion :- At constant pressure of the change H2O(s)→H2O(g)\mathrm{H}_2 \mathrm{O}(\mathrm{s}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g}) work done is negative.

Reason :- During phase transition work done is always negative. For the reaction, X2O4(l)⟶2XO2(g),ΔU=2.1 kcal,ΔS=20 cal K−1\mathrm{X}_2 \mathrm{O_4(l}) \longrightarrow 2 \mathrm{XO}_2(\mathrm{g}), \Delta U=2.1 \mathrm{~kcal}, \Delta S=20 \mathrm{~cal} \mathrm{~K}^{-1} at 300 K300 \mathrm{~K}. Hence, ΔG\Delta G is
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