test1s-1

# D δ s for a reaction between thermodynamic

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(d) Δ S for a reaction between thermodynamic equilibrium states approaches zero as T approaches zero, i.e. the entropy of a thermodynamic equilibrium state approaches a constant as T 0. (e) The vapor pressure of a compound is the equilibrium pressure of its vapor over a given condensed phase (solid or liquid) at a particular temperature. 2. This is a simple heat balance: q = 0 = heat of reaction + warming of cell = heat generated per hour × t + warming of cell = ( - 30 . 88 × 10 3 J / mol)(7 × 10 - 13 mol / h) t + (10 - 11 L)(1000 g / L)(4 . 2 J K - 1 g - 1 )(5 K) . t = (10 - 11 L)(1000 g / L)(4 . 2 J K - 1 g - 1 )(5 K) (30 . 88 × 10 3 J / mol)(7 × 10 - 13 mol / h) = 9 . 7 h . 3. (a) Δ r G m = - RT ln K = - (8 . 314 472 J K - 1 mol - 1 )(298 . 15 K) ln(0 . 087) = 6 . 05 kJ / mol . Δ r G m = Δ f G (aq) - Δ f G (g) Δ f G (aq) = Δ r G m + Δ f G (g) = 6 . 05 + ( - 33 . 33) kJ / mol = - 27 . 28 kJ / mol . 1

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(b) slope = - Δ r H m /R Δ r H = - R (slope) = - (8 . 314 472 J K - 1 mol - 1 )(2100 K) = - 17 . 46 kJ / mol . Δ r H m = Δ f H (aq) - Δ f H (g) Δ f H (aq) = Δ r H m + Δ f H (g) = - 17 . 46 + ( - 20 . 50) kJ / mol = - 37 . 96 kJ / mol . (c) Δ r G m = Δ r H m - T Δ r S m Δ r S m = Δ r H m - Δ r G m T = - 17 . 46 - 6 . 05 kJ / mol 298 . 15 K = - 78 . 9 J K - 1 mol - 1 . Δ r S m = S m (aq) - S m (g) S m (aq) = Δ r S m + S m (g) = - 78 . 9 + 205 . 77 J K - 1 mol - 1 = 126 . 9 J K - 1 mol - 1 . 4. (a) Δ f G = Δ f H - T Δ f S . By comparison, Δ f H = - 131 . 53 kJ / mol. (b) This is - Δ f S , where Δ f S is the standard entropy change of formation, i.e.
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