7/31/2023 0 Comments Equation maker chem![]() ![]() = equilibrium concentration of CO 2(g) in mol L -1 ![]() = equilibrium concentration of H 2(g) in mol L -1 = equilibrium concentration of CO (g) in mol L -1 = equilibrium concentration of H 2O (g) in mol L -1 (2) What is the relationship between what you know and what you need to find out? N(H 2(g)) (eq) = moles H 2(g) at equilibrium = 0.44 mol N(CO 2(g)) (i) = moles CO 2(g) initially = 1 mol N(H 2(g)) (i) = moles H 2(g) initially = 1 mol (1) What information (data) have you been given in the question? Insert these expressions for equilibrium concentrations into the R.I.C.E. concentration of C (g) will increase from 0 mol L -1 to (0 + c x) because C (g) is being produced by the decomposition of A (g).concentration of B (g) will increase from 0 mol L -1 to (0 + b x) because B (g) is being produced by the decomposition of A (g).concentration of A (g) will decrease from 1.0 mol L -1 to (1.0 - a x) because A (g) is being consumed during the reaction to produce products B (g) and C (g).In our example for the decomposition of A (g): Step 5: Use the initial concentration and concentration change expressions to a write an expression for the concentration of each species at equilibrium: Therefore the concentration of A must decrease by the amount ax (- a x), and, the concentrations of B and C must increase by bx (+ b x) and cx (+ c x) respectively. Note: since initially ONLY reactant A was present (no B, no C), clearly the reaction must proceed in the forward direction to produce products B and C. ![]()
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