Changing the volume of a reaction vessel affects the equilibrium position of a chemical reaction because it alters the concentrations of the reactants and products. According to Le Chatelier's principle, if a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the system will adjust its equilibrium position to counteract the change.For a gaseous reaction, decreasing the volume of the reaction vessel increases the pressure, which in turn affects the equilibrium position. The system will shift in the direction that reduces the pressure by favoring the side with fewer moles of gas.Let's consider the following reaction as an example:N2 g + 3H2 g 2NH3 g Initially, let's assume the following equilibrium concentrations at a certain volume V:[N2] = 0.50 M[H2] = 0.75 M[NH3] = 0.25 MNow, let's decrease the volume of the reaction vessel by a factor of 2 V/2 . To calculate the new equilibrium concentrations, we first need to determine the reaction quotient Q and compare it to the equilibrium constant K .Q = [NH3]^2 / [N2] * [H2]^3 Using the initial concentrations:Q = 0.25 ^2 / 0.50 * 0.75 ^3 = 0.0625 / 0.10547 0.593Assuming the reaction has an equilibrium constant K > Q, the reaction will shift towards the products to re-establish equilibrium. To calculate the new equilibrium concentrations, we can use an ICE Initial, Change, Equilibrium table:`` N2 + 3H2 2NH3Initial 0.50 0.75 0.25Change -x -3x +2xFinal 0.50-x 0.75-3x 0.25+2x``Since the volume is decreased by a factor of 2, the new equilibrium concentrations will be:[N2] = 2 0.50 - x [H2] = 2 0.75 - 3x [NH3] = 2 0.25 + 2x Now, we can use the equilibrium constant expression:K = [NH3]^2 / [N2] * [H2]^3 Substitute the new equilibrium concentrations:K = [ 2 0.25 + 2x ^2] / [ 2 0.50 - x * 2 0.75 - 3x ^3]To solve for x, we would need the value of K. However, without the value of K, we cannot determine the exact new equilibrium concentrations. But we can qualitatively say that the reaction will shift towards the products NH3 due to the decrease in volume, as it reduces the total number of moles of gas and thus reduces the pressure.