This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
ScanSolve AI Answer
Alright — let's do this. This question applies Le Chatelier's Principle, which states that if a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress. For pressure changes, this principle applies to reactions involving gases. An increase in pressure will shift the equilibrium towards the side with fewer moles of gas. Let's analyze each option: A) 2HgO(s) → 2Hg(l) + O_2(g) Reactant side: 0 moles of gas (solid) Product side: 1 mole of gas (O_2) An increase in pressure will shift the equilibrium to the left (reactant side), producing less* product. B) N_2(g) + 3H_2(g) 2NH_3(g) Reactant side: 1 mole of N_2 + 3 moles of H_2 = 4 moles of gas Product side: 2 moles of NH_3 = 2 moles of gas An increase in pressure will shift the equilibrium to the right (product side), which has fewer moles of gas. This will produce more* product (NH_3). C) CaCO_3(s) CaO(s) + CO_2(g) Reactant side: 0 moles of gas (solid) Product side: 1 mole of gas (CO_2) An increase in pressure will shift the equilibrium to the left (reactant side), producing less* product. D) H_2(g) + Cl_2(g) 2HCl(g) Reactant side: 1 mole of H_2 + 1 mole of Cl_2 = 2 moles of gas Product side: 2 moles of HCl = 2 moles of gas Since the number of moles of gas is equal on both sides, an increase in pressure will have no effect* on the position of the equilibrium. Therefore, an increase in pressure will produce more of the product in reaction B. The final answer is B) N_2(g) + 3H_2(g) 2NH_3(g). Send me the next one 📸