Question

The reaction CO_2(g) +C(s) doubleheadarrow 2CO(g) has K_p = 5.78 at 1200 K. Calculate the total pressure at equilibrium when 4.62 g of CO_2 is introduced into a 10.0-L container and heated to 1200 K in the presence of 3.73 g of graphite. Express your answer to three significant figures and include the appropriate units. P_total = Repeat the calculation of part A in the presence of 0.49 g of graphite. Express your answer to three significant figures and include the appropriate units. P_total =

Answer 1

When heated with graphite CO₂(g) forms CO(g)

CO₂(g) + C(s) $\rightarrow$ 2CO(g)

mass of CO₂ = 4.62 g

Molar mass of CO₂ = 44 g/mol

Number of moles of CO₂ = 4.62 g/44 g/mol = 0.105 mol

mass of C(s) = 3.73 g

molar mass of C(s) = 12 g/mol

number of moles of C(s) = 3.73 g/12 g/mol = 0.311 mol

therefore, C(s) is an excessive reactant.

Initial volume of CO₂(g), V = 10 L

Initial temperature of CO₂(g), T = 1200 K

R = 0.0821 L atm mol^-1 K^-1

Applying Ideal gas equation, PV = nRT

or, P = nRT/V = 0.105 mol 0.0821 L atm mol^-1 K^-1 1200 K / 10 L = 1.035 atm

  CO₂(g) + C(s)    $\rightleftharpoons$ 2CO(g)

Initial pressure 1.035 0

Equilibrium pressure (1.035-x) 2x

Kp = (Pressure of CO)²/(Pressure of CO₂) = 5.78 (given)

or, (2x)²/(1.035-x) = 5.78

or, 4x² + 5.78x - 5.9823 = 0

Solving the above quadratic equation, x = 0.7

Hence, at equilibrium P_CO2 = 1.035-x = 1.035 - 0.7 = 0.335 atm and P_CO = 2x = 2 X 0.7 = 1.4 atm

Hence, total preessure, P_total = P_CO2 + P_CO = 0.335 atm + 1.4 atm = 1.735 atm