At $1000^oC$, we have the following equilibrium:
$C(s)$ $+$ $2H_2(g)$ $CH_4(g)$,
with $K_p$ $=$ $0.263atm^{-1} $
$0.250 $ mol of $CH_4$ are injected in an empty container of $4$ L and it is waited for the equilibrium to be established.
Calculate the partial pressure of $CH_4$ at equilibrium.
Initially :
$p_{CH_4}$ $=$ $\frac{n_{CH_4}\cdot R\cdot T}{V}$ $=$ $\frac{0.250 \cdot 0.082\cdot 1073.15}{4}$ $=$ $5.50 $ atm
At a given temperature and a fixed volume, the partial pressures are proportional to the mole numbers.
With partial pressures it may therefore be reasoned as with moles:
$C(s)$
+
$2H_2(g)$
$CH_4(g)$
initially:
...
0 atm
5,50 atm
variation:
...
+2x atm
-x atm
at equilibrium:
...
2x atm
5,50-x atm
So we have:
$K_p$ $=$ $\frac{p_{CH_4}}{p_{H_2}^2}$
$0.263$ $=$ $\frac{5.50-x}{(2x)^2}$
$x$ $=$ $1.86$ or $x$ $=$ $-2.81$
(to be rejected, because the partial pressure of $CH_4$ cannot increase )
Therefore, at equilibrium:
$p_{CH_4}$ $=$ $5.50$ $-$ $1.86$ $=$ $3.64$ atm
$CH_4(g)$,
with $K_p$ $=$ $0.263atm^{-1} $
$0.250 $ mol of $CH_4$ are injected in an empty container of $4$ L and it is waited for the equilibrium to be established.
Calculate the partial pressure of $CH_4$ at equilibrium.