2.e. CH3OH(g)⇌CO(g)+2H2(g)
The reaction system represented by the equation is allowed to achieve equilibrium at a different temperature. The following table gives the partial pressure of each species in the equilibrium mixture.
Substance | Partial Pressure at Different Temperature |
---|
CH3OH(g) | 2.7 atm |
CO(g) | 4.2 atm |
H2(g) | 8.4 atm |
2.e.i. Use the information in the table to calculate the value of the equilibrium constant, Kp, at the new temperature.
Kp=PCH3OHPCOPH22=2.7(4.2)(8.4)2=110
2.e.ii. The volume of the container is rapidly doubled with no change in temperature. As equilibrium is re-established, does the number of moles of CH3OH(g) increase, decrease, or remain the same? Justify your answer by comparing the value of the reaction quotient, Q, with the value of the equilibrium constant, Kp.
The number of moles of CH3OH(g) will increase. When the volume is doubled, the partial pressures of all species are halved. The new value of Q will be: Q=1.35(2.1)(4.2)2=27.6. Since Q < Kp, the reaction will shift to the right, increasing the amount of products and decreasing the amount of reactants, so the number of moles of CH3OH(g) will increase.