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For the reaction CO(g)+2H(2)(g)hArrCH(...

For the reaction
`CO(g)+2H_(2)(g)hArrCH_(3)OH(g)`
Hydrogen gas is introduced into a five-litre flask at `327^(@)C`, containing `0.2` mol of `CO(g)` and a catalyst, untill the pressure is `4.92 atm`. At this point, `0.1` mol of `CH_(3)OH(g)` is formed. Calculate the equilibrium constants `K_(p)` and `K_(c )`.

Text Solution

Verified by Experts

Let the number of moles of hydrogen introduced be `n` moles.
Total moles of `CO` and hydrogen `=0.2+n`
Applying, `PV=nRT`
`P=4.92 "atm", V=5 L, R=0.082`
`T=(273+327)=600 K`
`4.92xx5=0.082xx600xx(0.2+n)`
or `0.2+n=(4.92xx5)/(0.082xx600)`
or `n=0.3 "mol"`
`CO(g)+2H_(2)(g)hArrCH_(3)OH(g)`
`{:(,0.2-x,,0.3-2x,,x,("Number of moles at equilibrium")),(or,0.2-0.1,,0.3-0.2,,0.1,),(or,0.1,,0.1,,0.1,):}`
`[CO]=0.1/5, [H_(2)]=0.1/5, [CH_(3)OH]=0.1/5" "("Active masses")`
Applying the law of mass action,
`K_(c)=([CH_(3)OH])/([CO][H_(2)]^(2))=(0.1/5)/(0.1/5 xx(0.1/5)^(2))=2500 "mol"^(-2) L^(2)`
We know that `K_(p)=K_(c)(RT)^(Deltan), Deltan=-2`
or `K_(p)=2500(0.082xx600)^(-2)`
or `K_(p)=2500/(49.2xx49.2)=1.0327 "atm"^(-2)`
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