Question

0.15 mol of CO taken in a 2.5 L flask is maintained at 705 K along with a catalyst so that the following reaction takes place CO(g) + 2$H_2$(g) $\rightleftharpoons$ $C{H}_3$OH(g). Hydrogen is introduced until the total pressure of the system is 8.5 atm at equilibrium and 0.08 mol of methanol is formed. $K_p$ and $K_c$ respectively are:

• A. = 0.045 and = 12.438
• B. = 150.85 and = 188
• C. = 0.045 and = 150.85
• D.

Answer 1

The correct option is C

= 0.045 and = 150.85

Initially there was 0.15 mol of CO (g) and at equilibrium 0.08 mol of CH₃OH was formed
$\begin{array}{cccccc}& CO\left(g\right)& +& 2H_2\left(g\right)& \leftrightharpoons & C{H}_{3}OH\\ Molesatstar& 0.15& & a& & 0\\ Molesatequb.& (0.15-x)& & (a-2x)& & 0.08\\ or& (0.15-0.08)& & (a-0.16)& & 0.08\end{array}$
Total moles at equilibrium $=(0.15-0.08)+(a-(2\times 0.08\left)\right)+0.08=a-0.01$
We can equate the above to n = number of moles at equilibrium from
n = $\frac{PV}{RT}$ P = 8.5 atm, R = 0.0821 atmL$K^{-1}$ $mo{l}^{-1}$, V = 2.5 litre and T = 705 K
n = 0.3676
No of moles of $H_2$ (g) at equilibrium = n - (0.15-0.08) - 0.08 = 0.2176
Now $K_c=\frac{\left[C{H}_{3}OH\right]}{\left[H_2{]}^2\right[CO]}$= 150.8585 $mol{e}^{-2}$$litr{e}^2$
$K_p$ much more straightforward once we figure out $K_c$.
$K_p$ = $K_c(RT{)}^{\Delta n}$

= 150.85 $\times$ $(0.0821\times 705{)}^{-2}$ = 0.045$at{m}^{-2}$