0.15 mole of CO taken in a 2.5L flask is maintained at 750 K along with a catalyst so that the following reaction can take place. CO(g)+2H2(g) →CH3OH(g) Hydrogen is introduced until the total pressure of the system is 8.5 atmosphere at equilibrium and 0.08 mole of methanol is formed. Calculate (a) KP and KC …

Q: 0.15 mole of CO taken in a 2.5L flask is maintained at 750 K along with a catalyst so that the following reaction can take place.
CO(g)+2H2(g) $ \rightleftharpoons $ CH3OH(g)
Hydrogen is introduced until the total pressure of the system is 8.5 atmosphere at equilibrium and 0.08 mole of methanol is formed. Calculate (a) KP and KC and
(b) the final pressure if the same amount of CO and H2 as before are used, but with no catalyst so that reaction does not take place.

Solution: CO(g) + 2H2(g) $ \rightleftharpoons $ CH3OH(g)
Initial (mole):  0.15            a       0
At eq. (mole):  (0.15–x)  (a–2x)    x
(0.15–0.08) (a–0.16) 0.08

But, x = 0.08 mole = 0.07

Total number of moles at equilibrium = a – 0.01

number of moles at equilibrium $= \frac{PV}{RT} = \frac{8.5 \times 2.5}{0.082 \times 750}$ = 0.35

∴ a– 0.01 = 0.35 ; a= 0.36

At equilibrium $[CO] = \frac{0.07}{2.5} $, $[H2] = \frac{0.02}{2.5} $ , $[CH3OH] =\frac{0.08}{2.5} $

$ K_C = \frac{\frac{0.08}{2.5}}{( \frac{0.07}{2.5}) ( \frac{0.02}{2.5})^2} $

KC = 178.57 M-2

KP = 178.57 (0.082 × 750)-2 = 0.047 atm-2

(b) When no reaction takes place, so total moles = 0.51

$P_{final} =\frac{n R T}{V} = \frac{0.51 \times 0.082 \times 750}{2.5}$ = 12.546 atm