122ch14c

122ch14c - 26 14.35 27 14.36 28 14.36(cont 29 14.37 14.40...

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Unformatted text preview: 26 14.35) 27 14.36) 28 14.36) (cont.) 29 14.37) 14.40) N2O5 (g) r = k [N2O5] v 2 NO2 (g) + 1/2 O2 (g) 1st order with respect to [N2O5], k = 6.82 x 10!3 s!1 at 70 °C Start with 0.0250 mol of N2O5(g) in 2.0-L container 0.0250 mol [N2O5]0 = --------------- = 0.0125 M = 0.012 M (rounding even) 2.0 L a) Use first-order integrated rate equation (three ways to write it): ln[A]t = !k t + ln[A]0 [A]t ln (-------) = !k t [A]0 [A]t = [A]0 e!k t !kt [N2O5]300s = [N2O5]0 e !3 !1 = (0.0125 M) e!(6.82 x 10 s )(300 s) = 1.6156 x 10!3 M moles = (1.6156 x 10!3 M)(2.0 L) = 3.23 x 10!3 M = 3.2 x 10!3 M 30 14.40) (cont.) b) Want time it takes for N2O5 to decrease to 0.010 mol 0.010 mol [N2O5]t = --------------- = 0.0050 M 2.0 L [N2O5]0 = 0.0125 M [N2O5]t ln (-----------) = !k t [N2O5]0 0.0050 M ln (--------------) = !(6.82 x 10!3 s!1) * t 0.0125 M t = 134.35 s = 2.24 min = 2.2 min c) 1st order, 0.693 0.693 t1/2 = -------- = -------------------- = 101.6 s = 102 s or 1.69 min k 6.82 x 10!3 s!1 14.43) Given a reaction, A(aq) ---> B(aq) and the data in problem 14.15 determine the order of the reaction (by making the appropriate graphs - integrated rate eqn), rate constant, and half-life. * continued on next page * 31 14.43) (cont.) 32 14.46) 33 14.48) a) In the reaction, H + Cl -----> HCl, because the reactants are atoms and spherical the orientation is less important. All collision orientations are equally effective. b) *** continued on next page *** 34 14.48) b) (cont.) 35 14.50) 14.52) 36 14.53) 37 14.53) (cont.) 38 14.55) k2 Ea 1 1 ln (-----) = ------ (---- - ----) k1 R T1 T2 or k1 Ea 1 1 ln (-----) = ------ (---- - ----) k2 R T2 T1 (Notes) (Book) The two eqns are equivalent (just look different). T in KELVIN, R = 8.314 J/molCK k1 = 2.75 x 10!2 s!1 T1 = 20 °C = 293 K Want the rate constant, k, 60 °C using two different activation energies, Ea. a) k2 = ? at T2 = 60 °C = 333 K when Ea = 75.5 kJ/mol Ea = 75.5 kJ/mol = 75.5 x 103 J/mol (units for Ea must agree with those of R) 75.5 x 103 J/mol 1 1 k2 ln (------) = --------------------- (--------- - ---------) k1 8.314 J/molCK 293 K 333 K k2 ln (------) = 3.7229 k1 (s.f. info is to the right of the decimal in a logarithm) k333 ------- = e3.7229 = 41.385 k293 k333 = k293 (41.385) = (2.75 x 10!2 s!1)(41.385) = 1.138 s!1) = 1.14 s!1 39 14.55) (cont.) b) k2 = ? at T2 = 60 °C = 333 K when Ea = 125 kJ/mol Ea = 125 kJ/mol = 125 x 103 J/mol (units for Ea must agree with those of R) k2 125 x 103 J/mol 1 1 ln (------) = --------------------- (--------- - ---------) k1 8.314 J/molCK 293 K 333 K k2 ln (------) = 6.1637 k1 (s.f. info is to the right of the decimal in a logarithm) k333 ------- = e6.1637 = 475.22 k293 k333 = k293 (475.22) = (2.75 x 10!2 s!1)(475.22) = 13.068 s!1) = 13.1 s!1 NOTE: k at 60 °C (333 K) is greater for the rxn with the larger Ea. k333, 125 kJ 13.1 s!1 -------------- = ----------- = 11.5 k333, 75.5 kJ 1.14 s!1 At 60 °C the k for Ea = 125 kJ/mol (higher Ea) is . 11.5 times greater than k for Ea = 75.5 kJ/mol. This tells us that an inc. in T has a greater effect for the rxn. with the large Ea. 40 14.57) 41 14.59) ...
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