4. Chemical bonding and molecular structure — Chemistry STD 11 Science — Question

$P : 10\ ml, 0.1\ M\ NaOH + 5\ ml, 0.1\ M\ HCl$

$Q : 10\ ml, 0.1\ M\ NaOH + 15\ ml, 0.1\ M\ CH_3COOH$

$R : 10\ ml, 0.1\ M\ NH_3 + 10\ ml, 0.1\ M\ NH_4Cl$

$S : 10\ ml, 0.05\ M\ NaF + 5\ ml, 0.1\ M\ HF$

Which of above solutions act as buffer

$\mathrm{X}_2(g) \rightleftharpoons 2 \mathrm{X}(g)$

The standard reaction Gibbs energy, $\Delta_r G^{\circ}$, of this reaction is positive. At the stiur of the reaction, there is one mole of $X_2$ and no $X$. As the reaction proceeds, the number of roles of $X$ formed is given by $\beta$. Thus, $\beta_{\text {equitibrium }}$ is the number of moles of $\mathrm{X}$ formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally. (Given : $R=0.083 \mathrm{~L} \mathrm{bar}^{-1} \mathrm{~mol}^{-1}$ )

($1$) The equilibrium constant $K_P$ for this reaction at $298 \mathrm{~K}$, in terms of $\beta_{\text {equilibs, um }}$, is

($A$) $\frac{8 \beta_{\text {equilibrium }}^2}{2-\beta_{\text {equilibrium }}}$    ($B$) $\frac{8 \beta_{\text {equititrium }}^2}{4-\beta_{\text {equilibrium }}^2}$     ($C$) $\frac{4 \beta_{\text {equilibrium }}^2}{2-\beta_{\text {equilibrium }}}$   ($D$) $\frac{4 \ell_{\text {equitibrium }}^2}{4-\beta_{\text {equilibrium }}^2}$

($2$) The $INCORRECT$ statement among the following, for this reaction, is

($A$) Decrease in the total pressure will result in formation of more moles of gaseous $\mathrm{X}$

($B$) At the start of the reaction, dissociation of gaseous $\mathrm{X}_2$ takes place spontaneously

($C$) $\beta_{\text {equilibrium }}=0.7$

($D$) $\quad K_c<1$

Given the answer question ($1$) and ($2$) 

Comment on optical activity of the products