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B(g)+C(g)hArrA(g). At constant temperatu...

`B(g)+C(g)hArrA(g)`. At constant temperature, mixture of 1 mol A(g), 2 mol B(g) and 3 mol C(g) are left to stand in a closed vessel of 1 L capacity. The equilibrium mixture is found to contain B(g) of 0.175 molar concentrations `(mol*L^(-1))`. Find the value of equilibrium constant at that temperature.

Text Solution

Verified by Experts

For the above reaction, `K_(c)=([A])/([B]xx[C])`. As given, at equilibrium `[B]=0.175mol*L^(-1)`. So, `2-0.175=1.825mol*L^(-1)` of B have participated in the reaction.
Now according to the equation, 1 mol B and 1 mol C combine together to form 1 mol A. so, 1.825 mol B and 1.825 mol of C combine together to produce 1.825 mol of A. thus at equilibrium at the concentration of A, B and C will be-
`{:("Initial conc. "(mol*L^(-1)):,B(g),+,C(g),hArr,A(g)),("Equilibrium conc. "(mol*L^(-1)):,0.175,,3-1.825=1.175,,1+1.825=2.825):}` [`because` Volume of the container=1L]
`therefore K_(c)=([A])/([B]xx[C])=(2.825)/(0.175xx1.175)=13.74`
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