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For , A + B hArr C. the equilibrium conc...

For , A + B `hArr` C. the equilibrium concentration of A and B at a temperäture are 15 mol `lit^(-1)` When volume is doubled the reaction has equilibrium concentration of A as 10 mol `lit^(-1)` then

A

`K_(c) = 2` mole `lit^(-)`

B

concentration of C in original equilibrium is 45 M

C

`K_(c) = 0.2 mol^(-1)` litre

D

as increasing volume the reaction proceeds in the backward direction

Text Solution

Verified by Experts

The correct Answer is:
B, C, D

Since, on increasing volume, pressure decreases and the reaction proceeds in the direction where it shows an increase in mole i.e backward reaction.
con. II equlibrium `[(15)/(2)+x][(15)/(2)+x][(a)/(2)-x]`
`:. (15)/(2) +x=10 :. x=(5)/(2)`
Now `Kc=([C])/([A][B])=([(a)/(2)-(5)/(2)])/([(15)/(2)+(15)/(2)][(15)/(2)+(15)/(2)])` ...(1)
`Kc=([C])/([A][B])=(a)/(15 xx 15)` ...(2)
Kc are same
`:. (((a-5))/(2))/(10 xx 10)=(a)/(15 xx 15) implies a=45M`
Now, `Kc=(45)/(15 xx 15)=0.2 "mole"^(-1)` lit
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