In reaction, CO(g) +2H_2(g) hArrCH_3OH(g)DeltaH=-92kJ//mol^(-1) Concentrations of hydrogen, carbon monoxide and methanol become constantat equilibrium . What will happen if (a) volume of the reaction vessel in which reactants and products are contained is suddenly reduced to half ? (b) partial pressure of hydrogen is suddenly doubled ? (c ) an inert gas is added to the system at constant pressure ? (d) the temperature is increased ?

In reaction, CO(g) +2H_2(g) hArrCH_3OH(g)DeltaH=-92kJ//mol^(-1) Conc

1.	In reaction, CO(g) +2H_2(g) hArrCH_3OH(g)DeltaH=-92kJ//mol^(-1) Concentrations of hydrogen, carbon monoxide and methanol become constantat equilibrium . What will happen if (a) volume of the reaction vessel in which reactants and products are contained is suddenly reduced to half ? (b) partial pressure of hydrogen is suddenly doubled ? (c ) an inert gas is added to the system at constant pressure ? (d) the temperature is increased ?
Answer» Solution :For the equilibrium, `CO(g) +2H_2 (g) hArr CH_3 OH(g)` `K_c=([CH_3OH])/([CO][H_2]^2)` `K_p=(P_(CH_3OH))/(P_(CO)xxP_(H_2)^2)` (a) When the volume of the vesselis suddenly reduced to half , the PARTIAL pressures of VARIOUS species gets doubled . Therefore, `Q_p=(2P_(CH_3OH))/(2P_(CO)xx(2P_(H_2))^2)=1/4K_p` SINCE `Q_P` is less than `K_p` , the equilibrium shifts in the forward direction producing more `CH_3OH`. (b) When partial pressure of hydrogen is suddenly doubled, `Q_p` changes and is no longer equal to `K_p`. `Q_p=(2P_(CH_3OH))/(2P_(CO)xx(2P_(H_2))^2)=1/4K_p` Equilibrium will shift from left to RIGHT. When and inert gas is added to the system at constant pressure, equilibrium shift from lower number of moles to higher number of moles(in backward direction). (d) By increasing the temperature, `K_p` will decrease and equilibrium will shift from right to left.