The equilibrium constant K_c for the decomposition of phosgene, COCl_2, is 4.63
ID: 1022501 • Letter: T
Question
The equilibrium constant K_c for the decomposition of phosgene, COCl_2, is 4.63 times 10^-3 at 527 degree C: COCl_2(g) CO(g) + cl_2(g) Calculate the equilibrium partial Pressure of all the components, starting with pure phosgene at 0.760 atm. Consider the following equilibrium process it 686 degree C: CO_2(g) + H_2(g) CO(g) + h_2O(g) The equilibrium concentrations of the reacting are [CO] = 0.050 M, (H_2) = 0.045 M, [CO_2] = 0.086 M, and [H_2O] = 0.040 M. (a) Calculate K_c for the reaction at 686 degree C. If we add CO_2 to increase its concentration to 0.50 mol/L. what will the concentrations of all the gases be when equilibrium is reestablished? Consider the heterogeneous equilibrium process: C(s) + C0O_2(g) 2CO(g) At 700 degree C, the total pressure of the system is found to be 4.50 atm. If the equilibrium constant K_P is 1.52, calculate the equilibrium partial pressures of CO_2 and CO. The equilibrium constant K_c for the reaction H_2(g) + CO_2(s) H_2O(g) + CO(g) is 4.2 at 1650 degree C. Initially 0.80 mol H_2 and 0.80 mol CO_2 are injected into a 5.0-L flask. Calculate the concentration of each species at equilibrium. Explain lessthanorequalto Chatelier's principle. How can this principle help us maximize the yields of reactions?Explanation / Answer
H2(g) + CO2(g) -------> H2O (g) + CO(g)
I 0.8 0.8 0 0
C -x -x +x +x
E 0.8-x 0.8-x +x +x
[H2] = 0.8-x/5
[CO2] = 0.8-x/5
[H2O] = x/5
[CO] = x/5
Kc = [H2O][CO]/[H2][Co2]
= x*x/(0.8-x)*(0.8-x)
4.2 =( x/0.8-x)2
2.05 = x/0.8-x
2.05*(0.8-x) = x
x = 0.54
[H2] = 0.8-x/5 = 0.8-0.54/5 = 0.052M
[CO2] = 0.8-x/5 = 0.8-0.54/5 = 0.052M
[H2O] = x/5 = 0.54/5 = 0.108M
[CO] = x/5 = 0.54/5 = 0.108M
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