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0.15 mole of CO taken in a 2.5L flask is...

0.15 mole of `CO` taken in a `2.5L` flask is maintained at `750K` along with a catalyst so that the following reaction can take place:
`CO(g)+2H_(2)(g)hArrCH_(3)OH(g)`
Hydrogen is introduced until the total pressure of the system is 8.5 atm at equilibrium and 0.08 mole of methanol is formed. Calculare (i) `K_(p)` and `K_(c)` and (ii) the final pressure if the same amount of `CO` and `H_(2)` as before are used, but with no catalyst so that the reaction does not take place.

Text Solution

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`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`
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