Question

Be sure to answer all parts. The equilibrium constant K, for the reaction Hz() + CO2() H2O(g) + CO(g) 1s 4.2 at 1650°C. Initially 0.76 mol H2 and 0.76 mol CO2 are injected into a 4.5-L flask. Calculate the concentration of each species at equilibrium. Equilibrium concentration of H2: Equilibrium concentration of CO2: Equilibrium concentration of H20: Equilibrium concentration of CO:

Answer 1

Given Hz (g) + CO2(g) = H2Ocg) + CO(g Kc = 4.2. We have , Heocg) + c( g H20) + CO2J) = 0.76 0.76 Initial nit mol change 27-2 - X tu ta Equil Ubrium 0.76-2 o 76-K . R. ...3

Kc = - [H2O ] [co] (H₂] [coz » Ke= coa xoa 10.76-) (0.76-2.] 22 0.5776-1.524ta² 422 - 1 - az 2 0.5 7 76 - 1.522 +22

3.242 6.384x + 2.1484259 0. Solving, we get a=104842, 0.51 x= 104842 is not acceptable as initial mole is 0.76. X 0.51 Moles & Equilibrium concentration Equillibrium moles' of H₂ = 0.76 -0. 51 = 0.25 mol of CO2 = 0.76~0.51 = 0.25 mo? Equillibrium moles of H₂O = 0:51 mol Equilibrium moles of co=0.5lmol. volume of container is 4.52, so molar concentration obtained by deviding each by 4.52. can be fquillibrium conontration 0.25 mol € 0.05 SM, LSL Equillibrium concentration of co ż=0-25 mol 0.055M. . 35M 45 Equil librium concentration of tho = 0.51 mol = 0.11 33M . 4.52 wibrium concentration of co-osl molao "2133 M. Equillibrium concentration o 4.522