$(R= 8.314\,\,JK^{-1}\,\,mol^{-1};\,\,ln\,2 = 0.693;\,\,ln\,3 = 1.098)$
Initially, Let $[{A_2}] = 1\,M$ and $[A] = 0\,M$
After $20\%$ dissociation , $80\%$ of $A_2$ remains.
$[{A_2}] = 1 \times \frac{{80}}{{100}} = 0.8\,M$
$20\%$ of $1\,M$ is
$1 \times \frac{{20}}{{100}} = 0.2.\,[A] = 2 \times 0.2 = 0.4\,M$
The equilibrium constant
$K = \frac{{{{[A]}^2}}}{{[{A_2}]}};$ $K = \frac{{{{[0.4]}^2}}}{{[0.8]}} = 0.2$
$\Delta {G^o} = - RT\,\ln \,K = - 8.314\,J{K^{ - 1}}\,mo{l^{ - 1}}$
$ \times 320\,K \times \ln \,0.2 = 4281\,J/mol$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(1)$ $\begin{array}{*{20}{c}}
{C{H_3}CH = C - C{H_3}} \\
{\,\,\,\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}}
\end{array}$
$(2)$ $C{H_3}CH = CHC{H_3}$
$(3)$ $\mathop {C{H_3}C}\limits^{\begin{subarray}{l}
\,\,\,\,\,\,\,\,{\begin{array}{*{20}{c}}
{}&H
\end{array}} \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,|
\end{subarray} } = CHC{H_2}C{H_3}$
$(4)$ $\begin{array}{*{20}{c}}
{C{H_3}C = C - C{H_3}} \\
{\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_{3\,\,}}\,\,C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$
| List $I$ (Mixture) | List $II$ (Separation Technique) |
| $A$ $CHCl _3+ C _6 H _5 NH _2$ | $I$ Steam distillation |
| $B$ $C _6 H _{14}+ C _5 H _{12}$ | $II$ Differential extraction |
| $C$ $C _6 H _5 NH _2+ H _2 O$ | $III$ Distillation |
| $D$ Organic compound in $H _2 O$ | $IV$ Fractional distillation |
Statement $I:$ In the titration between strong acid and weak base methyl orange is suitable as an indicator.
Statement $II:$ For titration of acetic acid with $\mathrm{NaOH}$ phenolphthalein is not a suitable indicator.
In the light of the above statements, choose the most appropriate answer from the options given below:
[Atomic number: $S =16, C 1=17, I =53$ and $Xe =54$ ]
