Question

52. A high temperatures carbon monoxide and hydrogen react to produce methanol, CO(g) + 2H2(g) CHOH(g) Δ}":-129 klinol Suppose that in an industrial reactor, the reaction is at equilibrium. For each of the following changes made to the equilibrium system a. The partial pressure of hydrogen is lowered, how will the partial pressures of b. Methanol is rapidly removed from the reactor, how will the pressures of the c· The temperature remains constant while the volume decreases, compressing d. The temperature is lowered. What happens to the methanol partial pressure? the other gases be affected as the reaction goes back to equilibrium? other gases be affected? the gas mixture. How will each of the gas partial pressures change? e. The reaction is slow at low temperature, but the system is still at equilibrium f. The temperature is raised, and pressure of carbon monoxide is lowered. How g. The total pressure is increased by compression, the temperature is lowered and A catalyst is added. How does this affect the partial pressure of methanol? is the partial pressure of hydrogen affected? the partial pressure of methanol is decreased. How would the partial pressure of carbon monoxide be affected?

Answer 1

The equilibrium constant for the reaction is given as

where the P terms denote partial pressures.

(a) When the partial pressure of H₂ is lowered, the denominator in the above expression decreases. However, K_p is an equilibrium constant and hence, must remain constant at a particular temperature. To keep K_p constant, the numerator must decrease proportionately, i.e, the reverse reaction is facilitated, producing more CO and H₂. Therefore, the partial pressure of CO must increase while the partial pressure of CH₃OH will fall.

(b) When methanol is rapidly removed from the reactor, the partial pressure of CH₃OH drops. To keep K_p constant, the equilibrium favors the forward reaction, producing more CH₃OH and depleting CO and H₂. Consequently, the partial pressures of CO and H₂ will fall.

(c) The reaction is attended by a decrease in the number of moles, i.e, Δn = -1. Assume the gases are ideal and hence, the volume is directly proportional to the number of moles. As the system is compressed, the volume decreases and hence, the system favors the side where there is a decrease in the number of moles of gases. Since the product side has fewer moles of gases, hence, the equilibrium favors the product. Consequently, the partial pressures of CH₃OH increases while those of CO and H₂ falls.