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NEET UG CHEMISTRY MCQ: THERMODYNAMICS

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Analyzing +4 | -1
Consider the following two reactions :

X→2Y;ΔHreaction =ΔH112X→Z;ΔHreaction =ΔH2\begin{aligned} & {X} \rightarrow 2 {Y} ; \Delta {H}_{\text {reaction }}=\Delta {H}_1 \\ & \frac{1}{2} {X} \rightarrow {Z} ; \Delta {H}_{\text {reaction }}=\Delta {H}_2 \end{aligned}

What would be the enthalpy change for the process Y→Z{Y} \rightarrow {Z} ?
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Correct option: A
ΔH2−12ΔH1\Delta {H}_2-\frac{1}{2} \Delta {H}_1
Option A (ΔH2−12ΔH1\Delta {H}_2-\frac{1}{2} \Delta {H}_1) is correct because it is obtained by manipulating the given reactions to arrive at the desired reaction.

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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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