Question

For the extremely important '"water gas shift" reaction shown below, the Kc= 6.74 x 10^-2 at a certain temperature. What are the equilibrium concentrations of all the components of the reaction if 0.742 moles of CO and H20 are initially mixed with 0.399 moles of CO2 and 0.941 moles of H2 in a 2.5L flask?

CO (g) + H20 <----> CO2 (g) + H2 (g)

Answer 1

CO (9) + H20 <---> CO2 (g) + H2 (9) Given that K. = 6.74 x 102 0.742 moles of CO and H2O 0.399 moles of CO2 and 0.941 Moles of H2 Volume =2.50 L molarity of CO = no of moles/volume in L = 0.742/2.50 = 0.2968M molarity of H2O = no of moles/volume in L = 0.742/2.50 = 0.2968M molarity of CO2 = no of moles/volume in L = 0.399/2.50 = 0.1596M molarity of H2 = no of moles/volume in L =0.941/2.50 = 0.3764M Now make the ICE table: ---------- CO(g) + H2O(g) ---------------> CO2(g) + H2(g) I --------- 0.2968 ---- 0.2968 ------------- 0.1596 ---- 0.3764 C ------- +x ------ + -------------- - ------ -X E ---- 0.2968+x ----- 0.2968+x -------- 0.1596-X ----- 0.3764-X kc = [CO2][H2]/[CO][H20] 6.74*10^-2 = (0.1596-x)(0.3764-x)/(0.2968+x)(0.2968+x) 6.74*10^-2*(0.2968+x)(0.2968+x) = (0.1596-x)(0.3764-x) X = 0.115 [CO] = 0.2968+x = 0.2968+0.115 = 0.4118 M [H20] = 0.2968+x = 0.2968+0.115 = 0.4118 M [CO2] = 0.1596-X = 0.1596-0.115 = 0.0446 M [H2] = 0.3764-X = 0.3764-0.115 = 0.2614 M Cross check kc = [CO2][H2]/[CO][H20] kc = [0.0446][ 0.2614][ 0.4118][ 0.4118] = 6.8*10^-2