Solutions - Practice Test - CHEM 112 Exam 1 1B. The rates of formation and decomposition are average rates that can be predicted by using the stoichiometry of the balanced equation to get: 1 D[ N 2O5 ] 1 D[ NO2 ] - = 2 Dt 4 Dt We can then solve for rate we re looking for: D [ N 2O5 ] 2 D[ NO2 ] 2-4 -4 Dt = - 4 Dt = - (5.5 10 4 ) = -2.75 10 The negative sign indicates that this is indeed a decomposition. 2C. For a first-order rxn the rate law looks like: r = k[a]. The correct units for this would be: M/s = k[m], and so k = M/Ms = 1/s.
3A. Since we re interested in the order of A we need to isolate A by finding two experiments where [A] is changing but [B] remains constant. Experiments 1 and 3 are good for this. We can then divide the rate law for experiment 1 by the rate law for experiment 3: r r 1 3 x 0.0585 k[0.2] [0.45] = = x 0.0585 k[0.5] [0.45] simplifies to æ 0.2 ö 1 = ç è 0.5 ø x = 0 x y y Thus the order is zero with respect to A, meaning that changing [A] has no effect on the rate of reaction. 4B. To find k we first will need to determine the order with respect to B using the same method as above. This time we need [A] to remain constant and [B] to change: r r 1 2 x 0.0585 k[0.2] [0.45] = = x 0.468 k[0.2] [0.90] y y 0.125 = (0.5) y y = 3 Now that we know what y is we can plug in the values for any experiment and solve for k: r1 = 0.0585 = k[0.2] 0 [0.45] 3 k = 0.642 5D. Straight up memorization. The y-axis needs to show 1/[A] and the slope must be positive.
6A. This is a slightly modified version of the half-life problem we did during the review. Notice that the mass of A starts at 80 g and after 20 mins has been reduced to half that value. Thus the halflife of A must be 20 mins. Next we just need the equation for half-life reactions: t1/2 = 0.693/k So k = 0.693/(20 mins) = 0.693/(1200 s) = 5.8 x 10-4 s -1 7B. Remember that we can manipulate the Arrhenius equation to contain the same value as the slope of this graph: k ln k ln k 2 1 2 æ 1 ç è T2 E a æ 1 1 ö = - R ç - T T è 2 1 ø - ln k1 E a = - 1 ö R - T 1 ø The left side of the equation is equal to the slope so we can substitute to get: - 3.97 10 E a 4 = 330kJ Ea = - R / mol
8E. The transition state is always at the top of the hill, where activation energy has been reached, so points B and D correspond to this. 9E. Since the second step is the slowest it represents the rate of the entire reaction which is given by: r = k2[d][b] However, we are not allowed to express the rate law in terms of an intermediate, which is what D is, so we must find a way to substitute D using the first step: [ D] K1 = [ A][ B] [ D] = K 1 [ A][ B] Now we can rewrite the rate for step 2: r = k2[d][b] = k2k1[a][b][b] = k2k1[a][b] 2 10E 30 hours is equal to exactly 5 half-lives, thus there will be 1/32 (equal to 0.03125) of the original amount left over (after 1 half-life: 1/2, after 2 half-lives: 1/4, after 3 three half-lives: 1/8, after four half-lives: 1/16, after 5 half-lives 1/32). 11C This question can be solved by just working out how many half-lives are required to achieve 75% decay. One half-life would cause 50% to decay, and the next would cause 25% to decay, which would result in 75% decay. Thus it takes 2 half-lives to achieve 75% decay. Since we re told that the entire process took 400 seconds and we now know that this time equals two halflives, we can divide by 2 to get 200 seconds as our half-life.
12B This problem is very similar to problem 7, though worded somewhat differently. To determine the percent still remaining we need to use the first order decay equation: æ N t ö ln ç = - kt N è 0 ø where the time is sixty-three years and all we need do is find the rate constant, k, and then we can solve for our answer. To determine k we must use the half-life equation: k = 0.693 28.8years = 0.0241-1 years Then, we can take the inverse natural log of the first order equation to get: æ N t ö -kt -(0.0241)(63) ç è N 0 = e ø = e = 0.219 = 21.9% 13B Since we re looking for the overall change in concentration over time, we can use the balanced equation to write the following: Δ [ N 2] ΔT = 1 Δ[ F 2 ] 3 ΔT = 1 2 We can then solve for D[NF3]/DT to get: Δ[ NF 3 ] ΔT = 2Δ [ N ] 2 = 2 ΔT 3 Δ[ NF 3 ] ΔT Δ F 2 [ ] ΔT Either answer would work, but only one option is given, B.
14D Since the reaction has only one step, its coefficients can be used as the exponents in the reaction s rate law. Therefore, the rate law is: Rate = k[a][b] 2. The units of rate are M/s and M for concentration, which leads to M -2 s -1 for the units of k. 15D By looking at experiments 1 and 2, we can see that tripling the concentration of [ClO2] causes the rate to increase by a factor of 9. Thus, the order is 2 with respect to [ClO2]. 16A To get the rate constant we first need to get the order with respect to hydroxide. By looking at experiments 2 and 3, we can see that tripling the hydroxide concentration causes the rate to increase by a factor of 3. Thus the order is 1 with respect to [OH]. At this point, you can take any experiment s data and plug in all the values that are now known. The only unknown variable is k, and so we can solve for it: #1: 0.0248 = k[0.060] 2 [0.030] 1 k = 229.6 17A A straight forward question. We can relate half-life to rate constant using the equation: t 12 = 0.693 k k = 0.00385/s
18D For a compound to be a intermediate, it must be produced during one of the first steps of the mechanism, and then completely used up in a later step. Thus intermediates never appear in the general reaction. As a rule, intermediates are first seen as products, and later as reactants. Only D fits this description. 19E This is conceptual question dealing with the application of the collision model theory. 20C For a first order reaction, the rate law would look as follows: M/s = k[m] Thus if we solve for k, we get: k = 1/s These are the correct units for every first order reaction. 21B The activation of the reverse reaction is the sum of the Ea of the forward reaction and the total change in energy with the opposite sign.
22C Considering that after one hour the radioisotope would have experienced a little under 3 halflives, we can expect the percentage remaining to be close to 12.5%. Remember that after one half-life 50% remains, after two half-lives 25% remains, and after three half-lives 12.5% remains. Since one hour doesn t quite make 3 half-lives, we expect there to be a little more left than 12.5%. Alternatively, you can use the first-order integrated rate law: ln [$] & = )*.,-./ [$] ' / 0 2 1 where: [N]t is the answer, [N]0 = 100% t = 60 mins t1/2 = 20.4 mins, 23C. Remember that in this situation spontaneous fusion is only likely to occur when the two nuclei are unstable. Since the question doesn t mention the stability of the nuclei, we have to assume that they are stable and so the reason they don t undergo fusion is due to the repulsion of the positively charged nuclei. Thus the answer is C. Note that answer choice A will never be the answer (catalysts) do not influence nuclear reactions), and that answer choices B, D, and E would likely apply to heavier nuclei (with 84 protons or more). 24C This is more memorization than anything else since neutron capture is not one of the reactions we see often. Refer to the section explaining Carbon Dating in the review guide to see how nitrogen bonds with a neutron due to cosmic radiation. However, if you simply balance material on both sides of the equation the only option that fits would be a particle with mass 1 amu and zero charge, thus a neutron.
25E The reason they have short lifetimes is because they react with electrons and produce high energy gamma rays, though this is not at all likely to appear on your test. 26E Below the belt of stability we are neutron deficient, and so a proton must be converted to a neutron to increase stability. This can be achieved by either positron emission or electron capture. 27E Hg has 80 protons and Au has 79 protons. Since the mass doesn t change we are simply looking for a reaction that reduces the proton number by 1. Either electron capture or positron emission would be acceptable answers, and so we pick electron capture as it s the only one of these two options available. 28A Only alpha decay can cause the mass to decrease by 4 in one step. 29D The quickest way to deal with questions like this is to follow the protons. We start with Calcium, which has 20 protons. First, electron capture reduces proton number by one, bringing us down to 19. Then, alpha decay reduces proton number by 2, bringing us down to 17. Thus the correct answer is chlorine since it has 17 protons. 30C Beta emission involves the loss of an electron. Therefore, carbon s proton number increases by one transforming it into N.
31C Since it is not stated, we can assume at this point in time there is Nt = 1 mg uranium present; therefore, there is 0.567 mg lead present in the rock. The question is then how much uranium was originally present in the rock if we assume all of the lead that exists comes from uranium decay. Since lead and uranium have different masses, the mass of uranium that decayed must be calculated by multiplying 0.567 mg Pb by 238 U/206 Pb (Mass number ratio), which gave a mass of 0.6551 mg Uranium. Therefore, the total amount of uranium originally present was 1.6551 mg, which is value for No. Plug this data into the first order rate decay integral equation and solve for t. 32E To get mass defect we subtract the sum of the 29 protons and 31 neutrons from the actual mass given: Dm = [(1.0072765 x 29) + (1.0086649 x 31)] [59.9338] = 0.5458 amu 33E Use the E=mc2 equation and work backwards. Solve for the mass defect (m), which will be in kilograms. Then, just convert to amu. 34A A nucleus that needs to undergo a proton-to-neutron conversion for stability is considered underneath the belt of stability. Both positron emission and electron capture are possible decays then.