MCQ
For a phase change, ${H_2}O(l)\underset {{0^{\,o}}C,\,1\,bar} \longleftrightarrow {H_2}O(s)$
- ✓$\Delta G = 0$
- B$\Delta S = 0$
- C$\Delta H = 0$
- D$\Delta U = 0$
✓
Answer
Correct option: A.
$\Delta G = 0$
a
$\Delta G = 0$ at equilibrium
$\Delta G = 0$ at equilibrium
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
In which of the following groups all the three members are of the alkaline earth metals family
The correct sequence of polarity of the following molecule
$(1)$ Benzene $(2)$ Inorganic Benzene
$(3)\, PCl_3F_2$ $(4)\, PCl_2F_3$
→$(1)$ Benzene $(2)$ Inorganic Benzene
$(3)\, PCl_3F_2$ $(4)\, PCl_2F_3$
$1\,\,\,$-$\,\,\,2\,\,\,$-$\,\,\,3\,\,\,$-$\,\,\,4$
(Where. $P =$ polar, $NP =$ non-polar)
The result of ozone hole is :
Consider the following cell reaction :
→$\mathrm{Cd}_{(s)}+\mathrm{Hg}_{2} \mathrm{SO}_{4(s)}+\frac{9}{5} \mathrm{H}_{2} \mathrm{O}_{(l)} \rightleftharpoons \mathrm{CdSO}_{4} \cdot \frac{9}{5} \mathrm{H}_{2} \mathrm{O}_{(s)}+2 \mathrm{Hg}_{(l)}$
The value of $\mathrm{E}_{\text {cell }}^{0}$ is $4.315\, \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. If $\Delta \mathrm{H}^{\circ}=-825.2\, \mathrm{~kJ} \,\mathrm{~mol}^{-1}$, the standard entropy change $\Delta \mathrm{S}^{\circ}$ in $\mathrm{J} \,\mathrm{K}^{-1}$ is ........ . (Nearest integer) [Given : Faraday constant $=96487\, \mathrm{C}\, \mathrm{mol}^{-1}$ ]
${K_{sp}}$ of an electrolyte $ AB$ is $1 \times {10^{ - 10}}.$ $[{A^ + }] = {10^{ - 5}}M,$ which concentration of ${B^ - }$ will not give precipitate of $AB$
→The ascending order of acidity of $-\mathrm{OH}$ group in the following compounds is :
→$(A)$ $\mathrm{Bu}-\mathrm{OH}$
$(B)$ $IMAGE$ $(C)$ $IMAGE$ $(D)$ $IMAGE$
Choose the correct answer from the options given below :
he atomic number is identical to the amount of $ ........ $ in the atom nucleus.
→Interstitial hydride can be formed by :
In the following sets of reactants which two sets best exhibit the amphoteric characters of $Al_2O_3. xH_2O$ ?
→Set $1$: $Al_2O_3 .xH_2O\, (s)$ and $OH^-(aq)$
Set $2$: $Al_2O_3 .xH_2O\, (s)$ and $H_2O\,(l)$
Set $3$: $Al_2O_3 .xH_2O\, (s)$ and $H^+(aq)$
Set $4$: $Al_2O_3 .xH_2O\, (s)$ and $NH_3(aq)$
Aluminium exist in $+3$ state where as thallium exist in both $+1\ \&+3$ oxidation state. Reason for this is
→