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For the system A(g)+2B(g) hArr C(g) the ...

For the system `A(g)+2B(g) hArr C(g)` the equilibrium concentration is
`A=0.06 mol L^(-1), B=0.12 mol L^(-1)`
`C=0.216 mol L^(-1)` The `K_(eq)` for the reaction is

A

250

B

416

C

`4xx10^(-3)`

D

125

Text Solution

Verified by Experts

The correct Answer is:
A
For reaction `A+2BhArrC`
`K=([C])/([A][B]^(2))=(0.216)/(0.06xx0.12xx0.12)=250.`
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