Question

$0.15$ mole of $CO$ taken in a $5L$ flask is maintained at $600K$ along with a catalyst, so that the following reaction takes place. $CO\left(g\right)+2H_2\left(g\right)C_6{H}_{5}CN\stackrel{\left(i\right)C_6{H}_{5}MgBr\left[excess\right],H_{2}O,H^{+}\left(ii\right)Zn-Hg,HCl}{\to }C{H}_{3}OH\left(g\right)$

Hydrogen is introduced until the total pressure of the system is $5.91$ atmosphere at equilibrium and $0.1$ mole of methanol is formed. Calculate $Kp$ and $Kc$. The final pressure if the same amount of $CO$ and ${}_2$ as before are used, but with no catalyst, so that the reaction does not take place.

Answer 1

$CO+2H_{2}O\to C{H}_{3}OH$

Mole before reaction 0.15 a 0

Mole after reaction 0.15-x a-2x x

Given x$=0.1$

Total moles$=0.15-x+a-2x+x=0.15-2x+a=a-0.05$ (Putting x=0.1)

Given pressure $P=5.91$atm

Volume $V=5$L

Temperature $T=600$K

According to ideal gas equation

$PV=nRT$

$n=\frac{5.91\times 5}{0.0821\times 600}=0.6$

So

$a-0.05=0.6$

$a=0.65$

Moles of $CO=0.15-0.1=0.05$

Moles of $H_2=0.65-2\times 0.1=0.45$

Moles of $C{H}_{3}OH=0.1$

$K_C=\frac{\left[C{H}_{3}OH\right]}{\left[H_2{]}^2\right[CO]}$

$=\frac{\frac{0.1}{5}}{\frac{0.45}{5}\times \frac{0.45}{5}\times \frac{0.05}{5}}$

$=250mo{l}^{-2}{L}^2$

$K_P=K_{C}R{T}^{\Delta n}$

$\Delta n=$moles of product$-$moles of reactant

$=1-3=-2$

$=250\times (0.0821\times 600{)}^{-2}$

$=0.1atm$