STD 11 - 1. Some basic concept of chemistry — JEE STD 11 Science — Question

$\left[\Lambda_{\mathrm{m}}=\right.$ molar conductivity

$\Lambda_{\mathrm{m}}^{\circ}=$ limiting molar conductivity

$\mathrm{c}=$ molar concentration

$\mathrm{K}_{\mathrm{a}}=$ dissociation constant of $\mathrm{HX}$ ]

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

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 ($2$) 

$A + B$ $\rightleftharpoons$ $C + D$ + heat

has reached equilibrium. The reaction may be made to proceed forward by