Chemical Equilibrium Question 8
Question 8 - 01 February - Shift 1
(i) $X(g) \rightarrow Y(g)+Z(g) K _{p 1}=3$
(ii) $A(g)$
$2 B(g)$ $K _{p 2}=1$
If the degree of dissociation and initial concentration of both the reactants $X(g)$ and $A(g)$ are equal, then
the ratio of the total pressure at $(\frac{p_1}{p_2})$ is equilibrium equal to $x: 1$. The value of $x$ is (Nearest integer)
Show Answer
Answer: (12)
Solution:
Formula: Degree of dissociation, calculation of $K_p$
$ x(g) \rightarrow y(g)+z(g) \quad k _{p_1}=3 $
Initial moles $n$
at equilibrium $n-\alpha$ an $\quad \alpha_n$
$ \begin{aligned} & k _{p_1}=\frac{(\frac{\alpha}{1+\alpha} \times p_1)^{2}}{\frac{1-\alpha}{1+\alpha} p_1} \\ & 3=\frac{\alpha^{2} \times p_1}{1-\alpha^{2}} \\ & A(g) \rightarrow 2 B(g) k _{p_2}=1 \end{aligned} $
Initial mole n
at equilibrium $x-\alpha n \quad 2 \alpha$
$p _{\text{total }}=p_2$
$k _{p_2}=\frac{(\frac{2 \alpha}{1+\alpha} \times p_2)^{2}}{\frac{1-\alpha}{1+\alpha} \times p_2}$
$1=\frac{4 \alpha^{2} \times p_2}{1-\alpha^{2}}$
$\frac{k _{p_1}}{k _{p_2}}=\frac{p_1}{4 p_2}$ $\frac{3}{1}=\frac{p_1}{4 p_2}$ $\therefore p_1: p_2=12: 1$ $x=12$
