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When 0.15 mol of CO taken in a 2.5 L fla...

When `0.15` mol of CO taken in a `2.5 L` flask is maintained at `750 K` along with a catalyst, the following reaction takes place
`CO(g)+2H_(2)(g) hArr CH_(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.
Calculate
a. `K_(p)` and `K_(c)`
b. 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

Verified by Experts

` {:((i),CO,+,2H_(2),hArr,CH_(3)OH) ,("Intialy :",0*15 "mole",,,,) ,("At eqm:",0*15 - 0*08 "mole",,,,), (,=0*017 "mole",,,,):}`
Total volume, `V= 2*5 L`
Total pressure ` P+ 8*5 "atm", T= 750 "K" `
` " Applying " PV= nRT,`
we get ` 8*5 xx 2*5 = n xx 0* 0821 * 750 or n = 0* 345 "mole" `
` :. "No. of moles of " H_(2) " at equilibrium " = 0* 345 - ( 0* 017 + 0*08 ) = 0* 248 " mol "`
` P_(CO) = (0* 017 )/(0*345 ) xx 8*5 "atm" = 0* 42 "atm" `
` p_(H_(2)) = ( 0*248 )/(0* 345) xx 8*5 "atm " = 6* 11 "atm "`
` P _(CH_(3)OH) = (0*8)/(0*345) xx 8*5 "atm" = 1* 97 "atm" `
` K_(p) = (P_(CH_(3)OH))/(P_(CO) xx P_(H_(2)))= (1*97 )/(0*42 xx (6* 11)^(2) )= 0* 1256 `
` K_(c) = ([CH_(3) OH])/([CO][H_(2)]^(2))= (0* 08 //2*5)/((0* 017 //2*5 )(0*248 //2*5)^(2)) = 478 *2 ` n(ii) No. of moles of `H_(2) " taken intially " = 0* 248 + 2 xx 0* 08 = 0* 308 `
No. of moles of CO taken intially `= 0*15`
Applying PV = nRt , ` P xx 2*5 = 0* 458 xx 0* 0821 xx 750 or P = 11* 28 "atm " ` .
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