Question

For this reaction ;
CO (g) + 2 H2
(g) CH3OH(g
1.00L reaction vessel contains 1.2mol CO(g), 1.5mol H2
and
2.0mol CH3OH(g) . How will Ptotal Change as the system
approaches equilibrium equal concentrations of CO and
CH3OH are found at equilibrium at constant T? (Kc = 14.5)
Hint
• Calculate Q first
• Compare it with Kc
• See what it favours (Product or reactant)
• Determine how the number of moles favoured specie relates to total
pressure.
12

Answer 1

$$K_p = K_c(RT)^{1-1-2} = K_c(RT)^{-2}$$

If $T = const = 298K$:

$$K_p = {14.5 \over (8.314*298)^2}$$

$$K_p = 1.964*10^{-5}$$

Then:

$$\Delta_rH = -RTlnK_p$$

$$\Delta_rH = -8.314*298*ln(1.964*10^{-5})$$

$$\Delta_rH = -8.314*298*(-10.84) = 26 856.88 J = 26.86 kJ$$

$$Q = -\Delta_rH = -26.86kJ$$

Endothermic reaction; at equal concentrations of CO and CH₃OH, the reagents (CO and H₂) are in excess.

$$[CO] = 1.6, [H_2] = 2.3, [CH_3OH] = 1.6$$

Accordingly, the equilibrium has shifted towards the formation of reagents. Then the partial pressure of the product decreased, and the partial pressures of the reactants, respectively, increased. Their contribution to overall pressure has also increased.