For the following reaction at $50^{\circ} \mathrm{C}$ and at 2 atm pressure,

$$ 2 \mathrm{~N}_2 \mathrm{O}_5(\mathrm{~g}) \rightleftharpoons 2 \mathrm{~N}_2 \mathrm{O}_4(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) $$

$\mathrm{N}_2 \mathrm{O}_5$ is $50 \%$ dissociated.

The magnitude of standard free energy change at this temperature is $x$.

$x=$ $\_\_\_\_$ $\mathrm{J} \mathrm{mol}^{-1}$ [Nearest integer].

Given : $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}, \log 2=0.30, \log 3=0.48, \ln 10=2.303$, ${ }^{\circ} \mathrm{C}+273=\mathrm{K}$

An electrochemical cell, consist of the following two redox couples, $\mathrm{M}^{x+} (\mathrm{aq}) / \mathrm{M}(\mathrm{s})\left[\mathrm{E}_{\text {red }}^{\Theta}=+0.15 \mathrm{~V}\right]$ and $\mathrm{Fe}^{3+}(\mathrm{aq}) / \mathrm{Fe}(\mathrm{s})\left[\mathrm{E}_{\text {red }}^{\Theta}=-0.036 \mathrm{~V}\right]$. The cell EMF $\left(\mathrm{E}_{\text {cell }}\right)$ is recorded to be 0.2057 V . If the reaction quotient of the electrochemical reaction is found to be $10^{-2}$, then the value of $x$ is

$\_\_\_\_$ .(Nearest integer)

[Given : M is a p-block metal and $\frac{2.303 R T}{F}=0.059 \mathrm{~V}$ ]

$$ \text { For a first order reaction } \mathrm{A} \rightarrow \mathrm{~B} $$

$$ \begin{array}{|l|l|} \hline \mathrm{t} / \min & {[\mathrm{A}] / \mathrm{M}} \\ \hline 0 & 0.6500 \\ \hline x & 0.0650 \\ \hline 20 & 0.00065 \\ \hline \end{array} $$

$x=$ $\_\_\_\_$ min. (Nearest integer)

In sulphur estimation, $2.0 \times 10^{-3} \mathrm{~mol}$ of an organic compound $(\mathrm{X})$ (molar mass $76 \mathrm{~g} \mathrm{~mol}^{-1}$ ) gave 0.4813 g of barium sulphate (molar mass $233 \mathrm{~g} \mathrm{~mol}^{-1}$ ). The percentage of sulphur in the compound $(\mathrm{X})$ is $\_\_\_\_$ $\times 10^{-1} \%$ (Nearest integer)