Identify the correct statement for change of Gibb’s energy for a system ($\Delta {G_{{\rm{system}}}}$) at constant temperature and pressure:

If $\Delta {G_{{\rm{system}}}}$=0, the system has attained equilibrium

If $\Delta {G_{{\rm{system}}}}$=0, the system is still moving in a particular direction

If $\Delta {G_{{\rm{system}}}}$=-ve, the process is not spontaneous

If $\Delta {G_{{\rm{system}}}}$=+ve, the process is spontaneous

JEE CHEMISTRY MCQ: THERMODYNAMICS

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Understanding +4 | -1

Identify the correct statement for change of Gibb’s energy for a system (

\Delta {G_{\mathrm{system}}}

) at constant temperature and pressure:

View answer and explanation

Correct option: D

\Delta {G_{\mathrm{system}}}

=0, the system has attained equilibrium

Option D is correct because at equilibrium, the change in Gibb's free energy is zero, indicating no further net change in the system.

Practice All Chapter MCQs →

\ \mathrm{C}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})-\mathrm{CO}(\mathrm{g}), \Delta \mathrm{H}_1=\mathrm{a}

\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})-\mathrm{CO}_2(\mathrm{~g}), \Delta \mathrm{H}_2=\mathrm{b}

\mathrm{C}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})-2 \mathrm{CO}(\mathrm{g}) ; \Delta \mathrm{H}_3=\mathrm{c}

0.18 \mathrm{~g}

4 \mathrm{~K}.

\left(\Delta_{\mathrm{C}} \text{H}^{\circ}=-2.8\times 10^6 \mathrm{~J} / \mathrm{mol} \text{ glucose}\right)

{CsI}

{CsCl}

{CsCl}

{Cs} I

\begin{array}{cccccccccccccccccccc} {}&{\Delta H(KJ/mol)} \\ {1/2A \to B}&{ + 150} \\ {3B \to 2C + D}&{ - 125} \\ {E + A \to 2D}&{ + 350} \end{array}

{B}+{D} \rightarrow {E}+2 {C}, \Delta {H}

\Delta H

\Delta u

\mathrm{C}{\mathrm{H}_4}\left( \mathrm{g} \right) + 2{\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right) + {\mathrm{H}_2}\mathrm{O}\left( l \right);\Delta H = - 890.3\mathrm{kJ}