InterviewSolution
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InterviewSolution
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`1` mol of A in `1` litre vessel maintained at constant T shows the equilibrium `A(g) hArr B(g) +2C(g) " " K_(C_(1))` `C(g) hArr 2D(g)+3B(g) " " K_(C_(2))` If the equilibrium pressure is `13/6` times of initial pressure and `[C]_(eq)=4/9[A]_(eq)`, Calculate `K_(C_(1))` and `K_(C_(2))`. |
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Answer» `{:(A(g),hArr,B(g)+,2C(g)and ,C,hArr,2D+,3B(g)),(1,,0,0,,,,),(1-a,,(a+3b),2a-b,2a-b,,2b,(a+3b)):}` `:. Sigma` mole at equilibrium `=1-a+a+3b+2a-b+2b` `=2a+4b+1` Now, `P_("initial") prop 1` `P_(eq) prop 2a+4b+1` `:. 2a+4b+1=13/6` or `2a+4b=7/6` Also `[C]_(eq)=4/9 [A]_(eq)` `2a-b=4/9xx(1-a)` `22a-9b=4` By equations (i) and (ii) `a=0.25, b=0.167` `:. K_(C_(1))=([B][C]^(2))/([A])=((a+3b)(2a-b)^(2))/((1-a))` `=([0.25+(3xx0.167)][0.5-0.167]^(2))/((1-0.25))=0.11` `:. K_(C_(2))=([D]^(2)[B]^(2))/([C])=((2b)^(2)(a+3b)^(3))/((2a-b))` `=([2xx0.167]^(2)[0.25+(3xx0167)]^(3))/([0.5-0.167])=0.142` |
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