Per Le Chatelier's principle, an increase in pressure on a gaseous system favors the side with fewer moles of gas. In the provided equilibrium reaction, the reactant side exhibits 4 moles of gas (1 mole CO and 3 moles H\(_2\)), while the product side has 2 moles of gas (1 mole CH\(_4\) and 1 mole H\(_2\)O). Consequently, an elevated pressure will displace the equilibrium to the right (forward direction), resulting in an augmented concentration of products and a diminished concentration of reactants.
- (A) The concentrations of reactants and products escalate due to the equilibrium shifting toward the products.
- (B) The equilibrium will advance in the forward direction, yielding increased quantities of CH\(_4\) and H\(_2\)O.
- (C) The equilibrium constant remains invariant, as pressure does not influence its value at a constant temperature.
Thus, the correct selection is (1) comprising only (A) and (B).
At a given temperature and pressure, the equilibrium constant values for the equilibria are given below:
$ 3A_2 + B_2 \rightleftharpoons 2A_3B, \, K_1 $
$ A_3B \rightleftharpoons \frac{3}{2}A_2 + \frac{1}{2}B_2, \, K_2 $
The relation between $ K_1 $ and $ K_2 $ is: