MCQ
Free energy change for the process $\ce{A(s)\leftrightarrow B(l)}$ will be :
- ✓$0$
- B$1$
- C$2$
- D$3$
✓
Answer
Correct option: A.
$0$
In the given process, $A$ and $B$ are in equilibrium, and we know that the free energy change for equilibrium reaction is zero.
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
At $1000\, K$ and $2$ $atm$ pressure , a gaseous mixture of $CO$ and $CO_2$ in equilibrium with solid carbon has $84\%$ $CO_{(g)}$ by mass. Calculate $K_p$ for the reaction ; $C_{(s)} + CO_{2(g)} \rightleftharpoons 2CO_{(g)}$ at this temperature .....$atm$
→The order if stability of carbocations is :
$\text{CH}_2=\text{CH}-\text{CH}^{\oplus}_2\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{I}$
$\text{CH}_3-\text{CH}_2-\text{CH}^{\oplus}_2\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{II}$
→$\text{CH}_2=\text{CH}-\text{CH}^{\oplus}_2\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{I}$
$\text{CH}_3-\text{CH}_2-\text{CH}^{\oplus}_2\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{II}$
The ionic radii $\mathop A\limits^o $ of $C^{4-}$ and $O^{2-}$ respectively are $2.60$ and $1.40$. The ionic radius of the isoelectronic ion $N^{3-}$ would be :-
→Halides of $\ce{III A}$ group elements predominantly act as :
→The $IUPAC$ name of picric acid is
→Which set hydridisation is correct for the following compounds
→$N{O_2}$, $S{F_4}$ , $PF_6^ - $
Molality of $0.8 \ \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$ solution (density $1.06 \mathrm{~g} \mathrm{~cm}^{-3}$ ) is________ $\times 10^{-3} m$.
→The total number of electron that takes part in forming bonds in ${N_2}$ is
→Nitroethane can exhibit one of the following kind of isomerism
What mass of $AgI$ will dissolve in $1.0\, L$ of its saturated solution ? [Given $: K_{sp}(AgI) = 1.0\times10^{-16} ; (At\, wt.\, Ag = 108; I=127)$
→