CH 102 1st Edition Lecture 11Outline of Current Lecture I. Chemical EquilibriumII. Significance of the equilibrium constantCurrent LectureI. Chemical Equilibrium-Chemical processes are driven to equilibrium by thermodynamicsΔG = Δ H - TΔSIf Δ G is negative, then the reaction goes in the forward direction, until Δ G is zeroIf Δ G is positive, then the reaction goes in the reverse direction, until Δ G is zeroAt equilibrium, ΔG = 0, the reaction does not stopThe rate of the forward reaction equals the rate of the reverse reactionThe concentration of reactants and products is given by the equilibrium expression, where Kc is the temperature dependent equilibrium constantsa. Nitrogen dioxide reacts with carbon monoxide to give carbon dioxide and nitric oxide.b. NO2 (g) + CO (g) → CO2 (g) + NO (g)i. The reaction was found to be zero order in [CO] and the rate only depended on the concentration of nitrogen dioxide. These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.ii.0 2 4 6 8 10 1200.20.4Time (hr)[NO2], Mc. What is the reaction order with respect to NO2 and what is the value of the rate constant?i.0 2 4 6 8 10 12-2.4-1.9-1.4f(x) = − 0.1x − 1.49R² = 0.97Time (hr)ln[NO2]ii.0 2 4 6 8 10 1202468101214f(x) = 0.75x + 4.01R² = 1Time (hr)1/[NO2]iii. What is the reaction order for NO2? – 2iv. What is the value of the rate constant? - 0.751d. Write the equilibrium expression for the following reactioni. 2 N2O5 (g) → 4 NO2 (g) + O2 (g) 1.ii. H2 (g) + I2 (g) → 2 HI (g)1.II. Significance of the equilibrium constanta. For K << 1i. the reverse reaction is favored, the forward reaction does not proceed very farb. For K ≈ 0 i. neither reaction (forward of reverse) is favored, the forward reaction proceeds about halfwayc. For K >> 1i. the forward reaction is favored, the forward reaction proceeds essentially to completiond. Given the equilibrium concentrations of reactants and product, calculate the equilibrium constant, Kc, for the following reactioni. H2 (g) + I2 (g) → 2 HI (g)1. At equilibrium2. [H2] = 0.11 M [I2] = 0.11 M [HI] = 0.78 M a.3. Equilibrium partial pressures4. H2 = 2.6 atm [I2] = 2.6 atm [HI] = 19.1
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