Chemistry 43 Problem Set Spring 08 Solutions. Derive an expression for the integrated rate law for a reaction obeying d[a] = k[a] /. [A] [A] 0 d[a] = k [A] / d[a] t [A] = / 0 k [A] / [A] [A] 0 { [A] / 0 [A] /}. Derive an expression for the integrated rate law for a reaction obeying d[a] = k[a] 3. d[a] [A] 3 = k [A] d[a] t [A] 0 [A] = 3 0 [A] [A] [A] 0 [A] [A] 0 k
3. Nitrogen pentoxide decomposes according to the reaction N O 5(g NO (g + O (g This reaction is found to be first order with respect to N O 5(g with a rate constant k = 4.8 0 4 s. Initially,.0 bar of pure N O 5(g are introduced into a reaction vessel. Calculate the total pressure in the vessel after 000. seconds. n N O 5 n NO n O initial n 0 0 0 change αn 0 αn 0 αn 0 / final n 0 ( α αn 0 αn 0 / P tot = n totrt V =.0 bar n tot = n N O 5 + n NO + n O = n 0 ( + 3α/ n N O 5 = n 0 ( α = n o e kt α = e kt n tot = n 0 [ + 3 ] ( ekt = n [ 0RT + 3 ] [ V ( ekt = P 0 + 3 ] ( ekt [ + 3 ] ( exp{(4.8 04 s (000. s} =.6 bar 4. Calculate the half life of nitrogen pentoxide using the data given in problem 3. t / = ln k = ln 4.8 0 4 s =.4 03 s 5. Derive expressions for the half life associated with the reactions given problems and in terms of the initial concentrations of A. At the half life [A] = [A] 0 / The for the half-order reaction or {[A 0 ] / ([A] 0 / / } /
t / = k {[A 0] / ([A] 0 / / } For the third-orde reaction ([A] 0 / [A] 0 / = 3 [A] 0 or t / = 3 k[a] 0 6. For the reaction at 508K HI (g /H (g + /I (g the half-life is 35 minutes when the initial HI pressure is 0.0 bar and 3.5 minutes when the pressure is.0 bar. Show that the reaction is second order, and determine the rate constant. For a second-order reaction t / = k[a] 0 = k[hi] 0 Does Or does 0=0, check. k = (t / (t / = [HI] 0, [HI] 0,? 35 3.5 =.? t / = k[hi] 0 3.5min bar = 0.074min bar 7. The recombination of gas-phase iodine atoms to form molecular gas-phase iodine obeys the second-order rate law I (g I (g d[i] = k[i]. At a temperature of 95.K it is found that if the initial atomic iodine concentration is.00 M, the half-life of the atomic iodine is 7. 0 s. If a flask of fixed volume 3
contains atomic iodine with initial concentration of 3.00 M at 95.K, calculate the time required for the concentration to fall to.00 M. [I] d[i] t [I] 0 [I] = k 0 8. The reaction obeys the rate law k = [I] 0 t / = [I] [I] 0 (7. 0 s(.00 M = 7.0 09 (M s.00 M 3.00 M = 7.0 09 M s t t = 9.5 0 s NOBr (g NO (g + Br (g d[nobr] = k[nobr] with rate constant at 98.K of k = 0.80 L mol s. If 0.00 moles of pure NOBr are introduced into a.00 L flask at 98.K, calculate the time required for the concentration of gas-phase Br to be 0.0050 mol L. [NOBr] d[nobr] t [NOBr] 0 [NOBr] = 0 [NOBr] [NOBr] 0 k 0.00 mol [NOBr] 0 =.00 L = 0.00 M If [Br ]=0.0050 M, then from the stoichiometry of the reaction [NOBr] = [NOBr] 0 [Br ] = 0.00 M (0.0050 M = 0.090 M 0.090 M 0.00 M = 0.80 M s t t =.4 s 4
9. The reaction CH 3 CH(OHCHCH H O + CH 3 CHCCH is found to be first order in CH 3 CH(OHCHCH with an activation energy of 4 kj mol. By what factor is the rate of this reaction increased if the temperature is raised from 5 C to 00 C? k = Ae Ea/RT ( k = exp E [ a ] k R T T = exp ( 4 03 J mol [ 8.344J mol K 373K ] 98K = 3. 0 8 0. What activation energy implies a reaction rate to be doubled for a 0 C temperature rise from 5 C to 35 C? ( k = exp E [ a ] k R T T ln k = E a k R E a = R ln T [ T T ] T = (8.344J mol K (ln (/98K /308K = 5.9kJ mol. The rate law for the thermal decomposition of acetaldehyde according to the process is given by CH 3 CHO (g CH 4(g + CO (g d[ch 3CHO] = k[ch 3 CHO] 3/. The rate constant for the decomposition of acetaldehyde is found to obey the Arrhenius equation in the temperature range 300K T 500K with a pre-exponential A = 3.67 0 M / s and activation energy E a = 99. kj mol. Calculate the half life of acetaldehyde at 400. K if the initial concentration of acetaldehyde is.00 M. ( k = Ae Ea/RT = (3.67 0 M / s 99000 J mol exp (8.344 J mol K = 3.79 0 5 M / (400. K 5
= [CH 3 CHO] [CH 3 CHO] 0 d[ch 3 CHO] t [CH 3 CHO] = k 3/ 0 ( [CH 3 CHO] / [CH 3 CHO] / 0 ( ([CH 3 CHO] 0 / / [CH 3 CHO] / 0 / t / = ( ([CH3 CHO] 0 / / [CH 3 CHO] / 0 k 3.79 0 5 M / s (/.00 M /.00 M =.54 0 4 s. In the gas phase the reaction of chlorine with chloroform to produce carbon tetrachloride and hydrogen chloride Cl (g + CHCl 3(g CCl 4(g + HCl (g is pseudo half order in chlorine when CHCl 3 is in great excess; i.e. d[cl ] = k[cl ] / With identical excess chloroform concentrations, measurement of the rate constant at 75 K and finds k = 4.0 0 M / s and similar measurements at 775 K find the rate constant to be 0.30 M / s. Assuming Arrhenius behavior calculate the half life of chlorine gas at 750K when the initial chlorine concentration is.5 M. ln k = E ( a k k B T T k(750 ln 0.040 M / s ( = 64 K ln 0.30 0.040 = E ( a k B 75 K 775 K 75 K 750 K d[cl ] [Cl [Cl ] = k ] d[cl ] t / [Cl ] 0 [Cl ] = / 0 E a k B = 64 K k(750 = 0. M / s {[Cl ] / 0 [Cl ] / } {[Cl ] / 0 [Cl ] / 0 / } / t / = { ( [Cl ] / 0 } ( = {(.5 M / } = 6.5 s k 0. M / s k 6