Question

The water-gas shift reaction plays a central role in the chemical methods for obtaining cleaner fuels from coal:

CO(g)+H2O(g) $\leftrightharpoons$ CO2(g)+H2(g)

A Study was made in which equilibrium was reached with [CO]=[H2]=[H2O]=.100 M and [CO2]=.400 M. After .300 mol of H2 is added to the 2-L container ad equilibrium is reestblished, what are the new concentrations of ALL of the components?

Answer 1

The chemical reaction is as follows: CO(g) + H,O(g)근 CO2 (g) + H, (g) Write the expression for the equilibrium constant for the reaction as shwon below [Col[H,O] Substitute the given concentrations in the above expression 0.400 ΜΊκ (0.100 M (0.100 M)x(0.100 M) Added moles of H, is 0.350 mol Calculate the concentration of H, for the added moles. Concentration of H- On the additon of H2, concnetration of the products will be increased. Thus, according to Le Chatlier's principle, the equilibrium of the reaction will be shifted to reactans side. Number of moles of H 0.350 mol Volume of the container 2.00 L 0.175 M

Write the ICE table for the equilibrium as shown below. Concentration CO(g) + H2O(g) 근 CO2(g) + H2(g) Initial: Change: Equilibrium: 0.100 +x 0.100+x 0.400 - x 0.275-x 0.1000.100 0.400 0.100 0.175-х Now write the expression for the equilibrium constant. (0.400 -x) (0.275 - ) (0.100 +x)(0.100+ x= 0.0435 M Now calculate new equilibrium concentrations of the components as shown below. New equilibrium concentration of CO-0,100 x 0.100 0,0435 0.143 M New equilibrium concentration of H、0-0.100 x 0.143 M New equilibrium concentration of CO.-0.400-x 0.400-0.0435-0.356 M New equilibrium concentration of H2O-0.275-0.0435-0.232 M