From the reaction, ${{\rm{P}}_{\left( {{\rm{white}}} \right)}} \to {{\rm{P}}_{\left( {{\rm{red}}} \right)}};$ ∆H= -18.4 kJ it following that :

Red P is readily formed from white P

White P is readily formed from red P

White P cannot be converted to red P

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NEET UG CHEMISTRY MCQ: THERMODYNAMICS

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From the reaction,

{\mathrm{P}_{\left( {\mathrm{white}} \right)}} \to {\mathrm{P}_{\left( {\mathrm{red}} \right)}};

∆H= -18.4 kJ it following that :

View answer and explanation

Correct option: D

White P can be converted into red P and red P is more stable

Option D (White P can be converted into red P and red P is more stable) is correct because the negative ∆H indicates that the conversion of white phosphorus to red phosphorus is exothermic, meaning red phosphorus is more stable.

More Questions from THERMODYNAMICS

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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

{T_i}

is the initial temperature and

{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

\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,

\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}

300 \mathrm{~K}

. Hence,

\Delta G