Help with chemistry homework. Equilibrium problemQuestion:For the equilibriumCO(g) + H2O(g) CO2(g) + H2(g)the equilibrium constant is 4.0 at 800 K. In an experiment, 1.0 mol of carbon monoxide, 2.0 mol of water vapor, 3.0 mol of carbon dioxide and 4.0 mol of hydrogen gas were introduced into an empty container with a volume of 20.0 dm3 Calculate the concentration of carbon dioxide in the container since equilibrium has been established at 800 K.I have put up a ICE table and gotten the starting concentrations[CO]i=0,05[H2O]i=0,1[CO2]i=0,15[H2]i=0,20I tried putting the concentrations at equilibrium C(eq) as[CO]=0,05-x[H2O]=0,10-x[CO2]=0,15+x[H2]=0,20+xSince thats how the equilibrium was written. The equilibrium constant K=4 so:4,0=(0,15+x)(0,20+x)/(0,05-x)(0,10-x)And got that x1= 0,3268 x2= - 0,0108The negative i cant use (?) and the positive one is too big to subtract from my starting concentrations. And when adding the x1 into the equilibrium equation i get K=9,74 so its wrong anyway. Is there something im missing in equilibrium/acid base reactions/gas reactions/how to work with concentration when different amount of substance of each component is added?
>>1488849initially, the ratio [CO2][H2]÷[CO][H2O] is 6, which is greater than 4. therefore, the reaction will go from right to left to reach the equilibriumyou can start over and switch the -x's and the +x's, or you can simply use the negative solution x2 = -0.0108
