Consider the equilibrium `H _(2) + I _(2) Harr 2HI.` Calculate the equilibrium constant of the reverse reaction when the equlibrium concentration of `H _(2), I _(2) and HI` are `1.14xx10^(-2),0.12 xx10^(-2) and 2.50 xx10^(-2) mol L ^(-1),` respectively

Options

A. `46.4`
B. `0.021`
C. `18.42`
D. `0.054`

Answer

B

Explanation

Given,
`H2 + I _(2) hArr 2HI`
Equilibrium concentration of `H_(2)` i.e.,
`[H_(2)]=1.14 xx 10 ^(-2) mol L ^(-1)`
`[I _(2)] =0.12 xx 10 ^(-2) mol L ^(-1)`
`[HI] = 2.52 xx 10 ^(-2) mol L ^(-1)`
The equation for eqilibrium constnat of the reverse is given by,
`K= ([H _(2)] [I_(2)])/([HI]^(2))`
`= (1.14 xx 10 ^(-2) xx 0.12 xx 10 ^(-2))/((2.52x 10 ^(-2))^(2))`
`= (0.1368 xx10^(-4))/( 6.3504xx10^(-4)) = 0.021`
`therefore K = 0.021`