Question

$2$ mole each of $S{O}_3,CO,S{O}_2$ and $C{O}_2$ is taken in a one lit. vessel. If $K_c$ for
$S{O}_3\left(g\right)+CO\left(g\right)\rightleftharpoons S{O}_2\left(g\right)+C{O}_2\left(g\right)$ is $1/9$ then:

• A. Total no. of moles at equilibrium are less than $8$
• B. $n\left(S{O}_3\right)+n\left(C{O}_2\right)=4$
• C. $\left[n\right(S{O}_2\left)/n\right(CO\left)\right]<1$
• D. Both $\left(B\right)$ and $\left(C\right)$

Answer 1

Answer: (D)

Both $\left(B\right)$ and $\left(C\right)$

$S{O}_3\left(g\right)+CO\left(g\right)\rightleftharpoons S{O}_2\left(g\right)+C{O}_2\left(g\right)\begin{array}{c}att=0\\ ateq\end{array}\begin{array}{c}2\\ 2-x\end{array}\begin{array}{c}2\\ 2-x\end{array}\begin{array}{c}2\\ 2+x\end{array}\begin{array}{c}2\\ 2+x\end{array}$

$K_c=\frac{\left[C{O}_2\right]\left[S{O}_2\right]}{\left[S{O}_3\right]\left[CO\right]}$

$\frac{1}{9}=\frac{{}^{n}C{O}_2{.}^{n}S{O}_2}{{}^{n}S{O}_3{.}^{n}CO}$

$\frac{1}{9}=\frac{(2+x{)}^2}{(2-x{)}^2}$

Using components & dividends

$\frac{1+9}{1-9}=\frac{(2+x{)}^2+(2-x{)}^2}{(2+x{)}^2-(2-x{)}^2}$

$=\frac{10}{-8}=\frac{8+2x^2}{8x}$

$40x=-32-8x^2$

$2x^2+10x+8=0$

$x^2+5x+4=0$

$(x+4)(x+1)=0$

$x=-4,-1$

reject

So, At eqm

$nS{O}_3=2-x=2-(-1)=3$

$nCO=3$

$nS{O}_2=1$

$nC{O}_2=1$

$\to$ Total no. of moles $=8$

$\to n\left(S{O}_3\right)+n\left(C{O}_2\right)=4$

$\to \frac{n\left(S{O}_2\right)}{n\left(CO\right)}=\frac{1}{3}<1$

$(\because 2+x=-2$ but moles cant be negative)