## Chemistry and Chemical Reactivity (9th Edition)

$n_{I_2,Reacted}=x, [I_2]_{Equil}=(0.105-x)/12.3$ $n_{I,Formed}=2x, [I]_{Equil}=2x/12.3$ Equilibrium constant: $K=[I]^2/[I_2]$ $3.76\times10^{-3}=(2x/12.3)^2/[(0.105-x)/12.3]$ $3.76\times10^{-3}\times(0.105-x)/12.3\times(12.3/2)^2=x^2$ $x^2=0.00121-0.0116x$ $x^2+0.0116x-0.00121=0$ $x=0.0295$ At equilibrium: $[I_2]=0.0061\ M, [I]=0.0048\ M$