1.

For the system 3A+2BhArrC, the expression for equilibrium constant is

Answer»

`([3A][2B])/(C)`
`([C])/([3A][2B])`
`([A]^(2)[B]^(2))/([C])`
`([C])/([3A]^(3)[2B]^(2))`

Solution :Equlibrium CONSTANT for the reaction, `3A+2BhArrC isK=([C])/([A]^(3)[B]^(2)).`


Discussion

No Comment Found

Related InterviewSolutions