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Reaction Feasibility

Reaction Feasibility. Will a specific reaction occur ?. How can we predict if a reaction is feasible ?. NaHCO 3 + HCl → NaCl + H 2 O + CO 2. Δ H is +ve (endothermic). Entropy increases. We know this reaction is feasible. How can we predict if a reaction is feasible ?.

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Reaction Feasibility

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  1. Reaction Feasibility Will a specific reaction occur ?

  2. How can we predict if a reaction is feasible ? NaHCO3 + HCl → NaCl + H2O + CO2 ΔH is +ve (endothermic) Entropy increases We know this reaction is feasible

  3. How can we predict if a reaction is feasible ? ΔG = ΔH - T ΔS ΔG = free energy ( kJmol-1) ΔH = enthalpy change ( kJmol-1) T = temperature (K) ΔS = entropy change ( kJK-1mol-1) ΔS must be converted from JK-1mol-1 (/1000)

  4. How can we predict if a reaction is feasible ? ΔG = ΔH - T ΔS A reaction is feasible if ΔG is negative ΔG < 0 A reaction becomes feasible when ΔG = 0

  5. Calculate ΔG for the formation of CO2 from graphite at 298K, givenΔHf = -393.5kJmol-1& ΔS is +3.3 JK-1mol-1 ΔG = ΔH - T ΔS (-393.5) – (298 x 3.3/1000) ΔG = -394.5 kJmol-1 Feasible reaction ΔG = -ve

  6. Is the decomposition of limestone spontaneous at standard temperature?ΔS = +165 JK-1mol-1 ΔH = +178 kJmol-1 ΔG = ΔH - T ΔS (+178) – (298 x 165/1000) ΔG = +128.83 kJmol-1 Not feasible ΔG = + ve

  7. Feasible  Spontaneous Thermodynamics tells us nothing about the speed a reaction happens It only tells us that it can happen C + O2 → CO2 ΔG = -394.5 kJmol-1 EA is large so reaction very slow at 298K

  8. Calculate ΔG for the rusting of Fe at 298K, givenΔHf = -825 kJmol-1& ΔS is -272 JK-1mol-1 2Fe(s) + 3/2O2(g) → Fe2O3 (s) ΔG = ΔH - T ΔS (-825) – (298 x (-272/1000) ΔG = -744 kJmol-1 Feasible reaction ΔG = -ve

  9. At what temperature does this reaction cease to be feasible?ΔHf = -825 kJmol-1 & ΔS is -272 JK-1mol-1 ΔG = ΔH - T ΔS Reaction becomes feasible when ΔG = 0 ΔH = T ΔS ΔH = T ΔS = 3033K T = -825/-0.272

  10. Free energy & Equilibrium ΔG = ΔH - T ΔS ΔG = 0 for a system at equilibrium Changes of state are systems at equilibria ΔS =ΔH T ΔH = T ΔS ΔH = T ΔS

  11. Calculate the entropy change when ice melts given thatΔHfusion (melting) = + 6.0 kJmol-1 ΔG = ΔH - T ΔS ΔG = 0 for a change of state ΔH = T ΔS ΔS =ΔH T ΔS = + 6/273 = 0.0219 kJK-1mol-1 = 0.0219 x 1000 JK-1mol-1 = 21.9 JK-1mol-1

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