Useful Formulas for Exam - General Chemistry | CHEM 152, Study notes of Chemistry

Download Useful Formulas for Exam - General Chemistry | CHEM 152 and more Study notes Chemistry in PDF only on Docsity! Useful Information ΔE = q+ w ΔE = nCvΔT ΔH = nCpΔT w = −PextΔV wrev = −nRT ln Vfinal Vinitial ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ΔS = dqrev T∫ S = k ln Ω( ) Ω = A! ai! i ∏ ΔS = nR ln Vfinal Vinitial ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ΔS = nCP ln Tfinal Tinitial ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ΔS = nCv ln Tfinal Tinitial ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ΔG = ΔG° + RT ln(Q) and K = Q (at equilibrium) aA + bB ⎯ → ⎯ ← ⎯ ⎯ cC + dD K = C[ ] c D[ ]d A[ ]a B[ ]b ΔH°rxn = cΔH° f products ∑ − cΔH° f reac tan ts ∑ ΔS°rxn = cS° products ∑ − cS° reac tan ts ∑ ΔG°rxn = cΔG° f products ∑ − cΔG° f reac tan ts ∑ ln(K ) = −ΔH° R 1 T ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ΔS° R ln K 2 K1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = −ΔH° R 1 T2 − 1 T1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Lecture 10: ΔG, Q, and K • Reading: Zumdahl 10.10, 10.11 • Outline – Relating ΔG to Q – Relating ΔG to K – The temperature dependence of K Relating ΔG to Q (cont.) • This equation tells us what the change in entropy will be for a change in concentration away from standard state. ΔSfi = ΔSfi°− R ln Pfinal Pinitial ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Entropy change for process occurring under standard conditions Additional term for change in concentration. (1 atm, 298 K) (P ≠ 1 atm) Relating ΔG to Q (cont.) • How does this relate to ΔG? ΔGrxn = ΔHrxn − TΔSrxn ΔGrxn = ΔH o rxn − TΔSorxn + RT ln Pfinal Pinitial ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ΔGrxn = ΔGrxn o + RT ln Pfinal Pinitial ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Relating ΔG to Q (cont.) ΔGrxn = ΔGrxn o + RT ln Pfinal Pinitial ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ΔGrxn = ΔGrxn o + RT ln Q( ) • Generalizing to a multicomponent reaction: Q = C[ ] C[ ]reference ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ coeff prod . ∏ C[ ] C[ ]reference ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ coeff react . ∏ • Where ΔG and K • The Reaction Quotient (Q) corresponds to a situation where the concentrations of reactants and products are not those at equilibrium. • At equilibrium, we have K. • What is the relationship between ΔG and K? ΔG and K (cont.) • At equilibrium, ΔGrxn = 0 ΔGrxn = ΔGrxn o + RT ln Q( ) 0 K 0 = ΔG°rxn +RTln(K) ΔG°rxn = -RTln(K) ΔG and K (cont.) • Let’s look at the interaction between ΔG° and K ΔG°rxn = -RTln(K) If ΔG° < 0 then Κ > 1 Products are favored over reactants reactants products ΔGrxn An Example • For the following reaction at 298 K: HBrO(aq) + H2O(l) BrO-(aq) + H3O+(aq) w/ Ka = 2.3 x 10-9 What is ΔG°rxn? ΔG°rxn = -RTln(K) = -RTln(2.3 x 10-9) = 49.3 kJ/mol An Example (cont.) • What is ΔGrxn when pH = 5, [BrO-] = 0.1 M, and [HBrO] = 0.2 M ? HBrO(aq) + H2O(l) BrO-(aq) + H3O+(aq) Q = H 3O +[ ] BrO−[ ] HBrO[ ] = 10−5( ) 0.1( ) (0.2) = 5x10−6 An Example (cont.) • Then: ΔGrxn = ΔGrxn o + RT ln Q( ) = 49.3 kJ/mol + (8.314 J/mol.K)(298 K)ln(5 x 10-6) = 19.1 kJ/mol ΔGrxn < ΔG°rxn “shifting” reaction towards products T Dependence of K (cont.) • Once we know the T dependence of K, we can predict K at another temperature: ln(K2 ) = −ΔH° R 1 T2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ΔS° R - ln(K1) = −ΔH° R 1 T1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ΔS° R ln K2 K1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = −ΔH° R 1 T2 − 1 T1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ the van’t Hoff equation. An Example • For the following reaction: CO(g) + 2H2(g) CH3OH(l) ΔG° = -29 kJ/mol What is K at 340 K? • First, what is Keq when T = 298 K? ΔG°rxn = -RTln(K) = -29 kJ/mol ln(K298) = (-29 kJ/mol) -(8.314 J/mol.K)(298K) = 11.7 K298 = 1.2 x 105 An Example (cont.) • Next, to use the van’t Hoff Eq., we need ΔH° CO(g) + 2H2(g) CH3OH(l) ΔHf°(CO(g)) = -110.5 kJ/mol ΔHf°(H2(g)) = 0 ΔHf°(CH3OH(l)) = -239 kJ/mol ΔH°rxn = ΣΔH°f (products) - ΣΔH° f (reactants) = ΔH°f(CH3OH(l)) - ΔH°f(CO(g)) = -239 kJ - (-110.5 kJ) = -128.5 kJ