`0.15 mol` of `CO` taken in a `2.5 L` flask is maintained at `750 K` alongwith a catalyst so that the following reaction can take place `CO(g)+2H_(2)(g) hArr CH_(3)OH(g)`. Hydrogen is introduced unit the total pressure of the system is 8.5 atm at equilibrium and 0.08 mol of methanol is formed. Calculate
a. `K_(p)` and `K_(c )`
b. The final pressure if the same amount of `CO` and `H_(2)` as brfore is used but no catalyst so that the reaction does not take place.

Explanation

`CO(g) + 2H_(2)(g) rarr CH_(3)OH`
Total moles of equilibrium = x - 0.1
`x - 0.1 = 8.5 xx 2.5 /(0.082 xx 750) = 0.34, x = 0.35`
(i) Kp = 0.056 , Kc = 213.33
(ii) Pressure = `nRT//V = [0.15+0.35]xx 0.0821 xx 750//2.5 = 12.315 atm`