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InterviewSolution
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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. |
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Answer» (`i`) Given, `{:(,,CO_((g)),+,2H_(2(g)),hArr,CH_(3)OH_((g))),(,"Mole before reaction",0.15,,a,,0),(,"Mole after reaction",(0.15-x),,(a-2x),,x):}` and `x=0.08` Total mole at equilibrium `=0.15-x+a-2x+x=0.15+a-2x` `=0.15+a-0.16=a-0.01` Also total mole at equilibrium are obtained by, `n=(PV)/(RT)` `n=(8.5xx2.5)/(0.0821xx750) :. P=8.5 atm, V=2.5 litre, T=750 K` at eqm. `:. n=0.345` `:. a-0.01=0.345` `:. a=0.355` At equilibrium, Mole of `CO=0.15-0.08=0.07` Mole of `H_(2)=0.355-0.16=0.195` Mole of `CH_(3)OH=0.08` `:. K_(c)=([CH_(3)OH])/([H_(2)]^(2)[CO])=(0.08//2.5)/(((0.07)/(2.5))((0.195)/(2.5))^(2))` `=187.85 mol^(-2)litre^(2)` Also `K_(p)=K_(c)(RT)^(Deltan)=187.85xx(0.0821xx750)^(2)` `=0.05 atm^(-2)` (`ii`) If reaction does not take place, then Mole of `CO=0.15` Mole of `H_(2)=0.355` `:. `Total mole `=0.505` `:. Pxx2.5=0.505xx0.0821xx750` `:. P=12.438 atm` |
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