14-2 Whether a reaction should proceed (thermodynamics) and how fast (kinetics) it should proceed.

Transcription

1 1-2 Whether a reaction should proceed (thermodynamics) and how fast (kinetics) it should proceed. 1-6 The rate of the reaction will slow down as time goes by since the rate is dependent upon the concentration of the reactants which are consumed during the reaction. 1-8 During the course of a reaction, the rate at which a reactant is consumed will decrease as amount of that reactant decreases The instantaneous rate of reaction is the infinitesimal change in reactant concentration over an infinitesimally short time period. rate = dx dt 1-11 For most reactions, the rate of reaction changes with time. The instantaneous rate of reaction is for a very short period of time. It can be measured more accurately than the changing rate of reaction, over a long period. 1-1 If the amount of NO 2 is measured in mol/liter, the units for the rate constant are M -1 time mol F 2 are depleted for each 1 mol Cl 2. Thus the rate of depletion of F 2 is three times that of Cl 2. Response (e). 1-2 d(i 2 ) dt = - d(mno ) dt x 5 mol I 2 2 molmno = -(-.56 x 10-3 M s -1 ) x 2.50 = M s For simple gas phase reactions, the rate law for the reaction is more likely to reflect the observed stoichiometry of the reaction A rate limiting step is the step in a multi-step kinetic mechanism which is the slowest and acts as the bottleneck in the reaction. 1-3 The first step and the slow, rate determining step are one and the same. Therefore, Rate = k(no) 2 (H 2 ) 1-35 The second-order rate equation suggests a slow, rate limiting step involving NO 2 + F 2, followed by a fast step for incorporation of the additional NO 2. Response (c) From the slow step (step 2): Rate = k (NO 2 )(NO 3 ) From step 1: K eq = [NO 2][NO 3 ] [N 2 O 5 ] [NO 2 ][NO 3 ] = K eq [N 2 O 5 ] Substituting into the rate expression, Rate = k K eq (N 2 O 5 ) 1-2 Thinking about collision theory, decreasing the temperature will decrease the number of collisions particles will undergo, and therefore decrease the zero order rate constant 1-7 If the volume is constant, the partial pressure of a compound is directly proportional to its concentration. Rate = k (NO) x (O 2 ) y From the data for the first two reactions we find that (100/150) x = 0.355/ The reaction is second-order in NO (x = 2). From the data for the last two reactions (130/180) = (1.0/1.). The reaction is first-order in oxygen. Rate = k (NO) 2 (O 2 ).

2 1-52 This is the integrated rate law for reactions which are first-order in a single reactant. Unimolecular dissociation and nuclear decay reactions are first-order reactions. Problems where one might want to determine the amount of material left after a long period of time are best solved with the integrated rate law For first-order N 2 O t 1-6 ln = - kt k = t 1/ 2 ln(0.057) / 5730 yr = 2. x 10 yr 1-68 A plot of the ln(n 2 O) versus time gives a straight line of slope -k equal to the rate constant. Thus, the reaction is first-order. Rate=k(N 2 O)

3 1-76 A plot of reactant concentration vs. time produces a linear plot. This means this is a zero order reaction and the slope of the plot is k, so k=0.015 sec A plot of the ln (Cr(NH 3 ) 5 Cl 2+ ) versus time gives a nearly straight line, indicating a possible firstorder reaction. The plot of 1/(Cr(NH 3 ) 5 Cl 2+ ) is not as linear. The reaction is most likely firstorder in (Cr(NH 3 ) 5 Cl 2+ ).

4 1-9 Assume that the activation energy remains essentially constant for small changes in temperature. k Ea 1 1 k = R T2 1 T1 2 e ln k1 k2 = E a 1 1 R T 2 T ln = Ea 1.53 x Ea = ( 2.87)(8.31).62x = 5.16 x 10 J/mol rxn = 51.6 kj/mol rxn A longer capillary should change the rate constant to a lower value but the process should remain first-order. 3.5 lnv time(s) A plot of ln V versus time produces a straight line, with a slope -k = -5.7 x 10-3 ml s -1. The reaction is first-order with a rate constant of 5.7 x 10-3 ml s lnv time(s) Addition of catalyst will affect the kinetics of the forward and reverse reactions, but not the thermodynamics of these reactions, therefore (b), (c), (d), and (e) will be affected by a catalyst. The others are thermodynamic functions.

5 1-111 (a) Rate = -k (H 2 O 2 ), because the first step is the rate determining step. The reaction is therefore first-order with respect to (H 2 O 2 ). (b) The data also indicate first-order with respect to (H 2 O 2 ). The t 1/2 = 835 sec. After 1670 sec (2 x t 1/2 ) only one fourth of the reactant is present. This matches the relationship for a first-order reaction given in Table 1.. (c) The t 1/2 = 835 sec. See (b). (d) The activation energy in the reverse reaction is larger since the starting point will be lower in energy for the products than for the reactants. (e)the turnip (somehow) is acting as a catalyst for the reaction. By increasing the rate by 600 x The Ea will be reduced by ln (600 x 10 9 ) or a factor of 27. Problem a.rate = -k[h2o2] b. time Conc time Ln[C] time 1/[Conc] b. Yes, first order reaction

6 c. Yes 835 sec d. Reverse e. Activation energy is significantly decreased.