I took the time to answer, so please rate, thanks! Using a ICE table, H2 + I2 <--> 2HI (Kc=2.5) Initial/M | 0.2 | 0.4 | 0.5 | Change | +x | +x | -2x | Equilibrium/M |0.2+x|0.4+x|0.5-2x| Kc=(0.5-2x)^2/(0.2+x)(0.4+x)= 2.5, we assume x~0. so (0.5-2x)^2=2.5(0.2)(0.4) x=0.0264 or 0.474(rejected) The concentration for [H2], [I2] and [HI] is 0.2264 M, 0.4264 M and 0.4472 M. (Check your required sig. figs. just in case) Hope that helped! Solution I took the time to answer, so please rate, thanks! Using a ICE table, H2 + I2 <--> 2HI (Kc=2.5) Initial/M | 0.2 | 0.4 | 0.5 | Change | +x | +x | -2x | Equilibrium/M |0.2+x|0.4+x|0.5-2x| Kc=(0.5-2x)^2/(0.2+x)(0.4+x)= 2.5, we assume x~0. so (0.5-2x)^2=2.5(0.2)(0.4) x=0.0264 or 0.474(rejected) The concentration for [H2], [I2] and [HI] is 0.2264 M, 0.4264 M and 0.4472 M. (Check your required sig. figs. just in case) Hope that helped!.