Question

The reaction CO2(g) + C(s)<-->2CO(g) has Kp=5.78 at 1200 K.

1) Calculate the total pressure at equilibrium when 4.73g of CO2 is introduced into a 10.0-L container and heated to 1200 K in the presence of 3.75g of graphite. (I have the answer to this half 1.77 atm)

2) Repeat the calculation of part A in the presence of 0.49g of graphite.

Answer 1

CO_2 (g) + C (s) doubleheadarrow 2 CO (g) k_p = 5.78 at 1200 k k_p = k_c (R^)^Delta n Delta n = Molar of product - Moles of reactants Delta n = 2 - 2 = 0 Therefore k_p = k_c (RT)^0 k_p = k_c \r\n k_c = [CO]^2/[C] [CO_2] Concentration unit = moles/liter given weight of CO_2 = 4.73 g given weight of C = 3.75 g C CO_2 = 4.73/44 times 10 = 0.01075 44 = molecular weight of CO_2 10 = volume \r\n C c = 3.75/12 times 10 = 0.03125 k_c = [CO]^2/0.01075 times 0.03125 \r\n [CO]^2 = 1.94172 times 10^- 3 [CO] = 0.044065 moles/liter 1 liter contains 0.044065 moles Therefore 10 liter contains 0.44065 moles Applying ideal gas equation PV = n RT \r\n n = n_CO_2 + n_CO + n_C } moles in 10 liters of volume n = 0.1075 + 0.44065 + 0.3125 n = 0.86065 R = 0.082057 L atm
sorry to say, but the answer that you mentioned is not correct. I have solved the problem, hope you understand it. Thank you!