In a reaction `A+2B hArr 2C, ` 2.0 moles of `'A'` 3 moles of 'B' and 2.0 moles of 'C' are placed in a 2.0 L flask and the equilibrium concentration of 'C' is 0.5 mol`//` L . The equilibrium constant (K) for the reaction is

Options

A. `0.073`
B. `0.147`
C. `0.05`
D. `0.026`

Answer

C

Explanation

` {:(,A,+,2B,hArr,2C),("Intial conc.",1 "mol "L^(-1),,1.5 " mol "L^(-1),,1 " mol "L^(-1)),("At eqm.",(1+0.25),,(1.5 + 0.5),,0.5),(,=1.25,,=2.0,," mol "L^(-1)),(," mol"L^(-1),," mol"L^(-1),,):}`
`K= (0.5)^(2)/(1.25 xx (2.0)^(2)) = 0.05`
Note that 0.5 mol `L^(-1)` of C has decomposed to produce 0.25 mol of A and .5 mol of B.