MCQ
The formula of a gaseous hydrocarbon which requires $6$ times of its own volume of $O _{2}$ for complete oxidation and produces $4$ times its own volume of $CO _{2}$ is $C _{ x } H _{ y }$. The value of $y$ is ...... .
- A$13$
- B$10$
- ✓$8$
- D$5$
$C _{ x } H _{ y ( g )}+\left( x +\frac{ y }{4}\right) O _{2}( g ) \rightarrow x CO _{2}( g )+\frac{ y }{2} H _{2} O (\ell)$
$V\,\quad\quad\quad\quad\quad6V\quad\quad\quad\quad\quad\quad-$
$-\,\quad\quad\quad\quad\quad-\quad\quad\quad\quad\quad Vx=4V$
$\Rightarrow x=4$
Sinc: $(I)$ $Vo _{2}=6 \times V _{ c _{ x } H _{y}}$
$\Rightarrow V \left( x +\frac{ y }{4}\right)=6 V$
$\left.\Rightarrow\left( x +\frac{ y }{4}\right)=6\right] \Rightarrow 4+\frac{ y }{4}=6$
$\Rightarrow y =8$
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$\mathrm{X}_2(g) \rightleftharpoons 2 \mathrm{X}(g)$
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 ($2$)