Question

The equilibrium pressure for the reaction $MS{O}_4\cdot 2H_{2}O\left(s\right)\rightleftharpoons MS{O}_4\left(s\right)+2H_{2}O\left(g\right)$ is $\pi /4$ atm at $400K$. The $K_p$ for the given reaction (in ${atm}^2$) is:

• A. ${\pi }^2/4$
• B. $\pi /6$
• C. ${\pi }^2/16$
• D. $\pi /16$
• E. $16/\pi$

Answer 1

The correct option is C ${\pi }^2/16$

$K_p=\frac{{\left[Partialpressureofproduct\right]}^x}{{\left[Partialpressureofreactant\right]}^y}=\frac{{\left(p\right)}_{product}^x}{{\left(p\right)}_{reactant}^y}$
(Equilibrium pressure)
($x$ and $y$ are powers over gaseous products and reactants respectively)
$\therefore K_p=\frac{{\left(P\right)}_{product}^2}{{\left(P\right)}_{reactant}^o}={\left(P\right)}_{Product}^2={\left(\frac{\pi }{4}\right)}^2=\frac{{\pi }^2}{16}{atm}^2$
Therefore, $K_p$ for given reaction
$\left(in{atm}^2\right)=\frac{{\pi }^2}{16}$