Carbon monoxide gas may be formed as follows. CO_2(g) + C(s) 2 CO(g) K_p = 5.78
ID: 958638 • Letter: C
Question
Carbon monoxide gas may be formed as follows. CO_2(g) + C(s) 2 CO(g) K_p = 5.78 at 1200 K 3.46g of CO_2, 1.98 g of C(s) and 7.22 g of CO(g) are placed in a 12.2 L container and heated to 1200 K. a) Calculate the total pressure (in bar) at equilibrium. All mathematical steps must be shown. b) If the volume of the vessel is allowed to expand to 40.0 L when the system is already at equilibrium, would you predict that the moles of CO(g) would increase, decrease or stay the same? Explain. c) If the system is at equilibrium and the temperature is decreased to 800 K, would you predict that the moles of CO(g) would increase, decrease or stay the same? Explain.Explanation / Answer
CO2(g) + C(s) <---> 2CO(g) ; Kp = 5.78
Moles of CO2 = mass/molar mass = 3.46/44 = 0.079
Partial pressure of CO2 initially = (n*R*T)/V = (0.079*0.0821*1200)/12.2 = 0.638
Moles of CO = mass/molar mass = 7.22/28 = 0.258
Partial pressure of CO initially = (0.258*0.0821*1200)/12.2 = 2.08
Now, reaction Quotient,Kq = PCO2/PCO2 = 6.796
Since Kq > Kp , reaction will proceed in the backward direction i.e CO2 formation will take place
Let at eqb., ,PCO2 = (0.638+x) & PCO = (2.08-2x)
Thus, Kp = (2.08-2x)2/(0.638+x)
or, 5.78*(0.638+x) = 4x2 -8.32x + 4.33
or, 3.69 + 5.78x = 4x2 -8.32x + 4.33
or, x = 0.046 atm
Thus, total pressure at eqb. = (2.08-2x) + (0.638+x) = 2.672 bar
b) since the number of moles in the product side is more than that in the reactant side, and increase in volume leads to decrease in partial pressure.Therefore to counteract that, the equilibrium will shift in the forward direction and hence CO formation will begin.
Thus, after attaining the equilibrium again, CO moles will increase.
c) Temperature directly effects the equilibrium constant value.Since the reaction is an endothermic reaction, decrease in temperature will decrease the Kc value and thus, moles of CO will decrease