Question

In a closed system: $A\left(s\right)\rightleftharpoons 2B\left(g\right)+3C\left(g\right),$ if the partial pressure of $C$ is doubled, then partial pressure of $B$ will be $\frac{x}{y\sqrt{2}}$ times the original value. Find the value of $x+y$

Answer 1

For a closed system-
$A\left(s\right)\rightleftharpoons 2B\left(g\right)+3C\left(g\right)$
$K_p=(P_B{)}^2\times (P_C{)}^3$
$\Rightarrow \frac{K_p}{(P_C{)}^3}=(P_B{)}^2$ ....(1)
Since, $K_p$ is constant, thus if the partial pressure of $C$ is doubled then $P{c}^{\prime }=2Pc.$
$\therefore K_p=(P_B^{\prime }{)}^2\times (P_C^{\prime }{)}^3$
$K_p=(P_B^{\prime }{)}^2\times (2P_C{)}^3$
$\Rightarrow \frac{K_p}{(P_C{)}^3}=8\times (P_B^{\prime }{)}^2$ ...(2)
From $e{q}^n$ (1) and (2), we have
$8\times (P_B^{\prime }{)}^2=(P_B{)}^2$
$\Rightarrow (P_B^{\prime }{)}^2=\frac{1}{8}\times (P_B{)}^2$
$\Rightarrow P_B^{\prime }=\frac{1}{2\sqrt{2}}\times P_B$
Hence, if the partial pressure of C is doubled then partial pressure of B will be $\frac{1}{2\sqrt{2}}$ times its original pressure.
Hence, $x=1$ and $y=2$, $x+y=2+1=3$.