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Thermal Decompostion Chapters' Solution, Exercises of Chemistry

University of Connecticut (UConn) - StamfordChemistry

Problem #5 is about the thermal decomposition of phosphine into phosphorus and hydrogen

Typology: Exercises

2021/2022

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Problems - Chapter 19 (with solutions)
1) Define the following terms:
a) catalyst - A substance that speeds up the rate of a reaction without itself being
consumed or produced.
b) half-life - The time it takes for the initial concentration of a reactant to decrease
to half (50 %) of its original value.
c) reaction intermediate - A substance that is produced in one step of a reaction
mechanism and which is consumed in another step in the mechanism. Reactants,
products, and catalysts are not reaction intermediates.
2) (19.10) Write the reaction rate expressions for the following reactions in terms of the
disappearance of the reactants and the appearance of products.
a) 2 H2(g) + O2(g) 2 H2O(g)
rate = - 1/2 ([H2]/t) = - ([O2]/t = 1/2 ([H2O]/t)
b) 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)
rate = - 1/4 ([NH3]/t) = - 1/5 ([O2]/t) = 1/4 ([NO]/t) = 1/6 ([H2O]/t)
3) Consider the reaction
2 N2O(g) 2 N2(g) + O2(g)
a) Express the rate of reaction with respect to each of the reactants and products.
rate = - 1/2 ([N2O]/t) = 1/2 ([N2]/t) = [O2]/t
b) In the first 15.0 s of the reaction, 0.015 mol of O2 is produced in a reaction
vessel with a volume of 0.500 L. What is the average rate of the reaction over this time
interval (including correct units)?
The concentration of O2 produced is
[O2] = 0.015 mol = 0.030 mol/L
0.500 L
average rate = [O2] =0.030 mol/L = 0.0020 mol/Ls
t 15.0 s
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Problems - Chapter 19 (with solutions)

  1. Define the following terms: a) catalyst - A substance that speeds up the rate of a reaction without itself being consumed or produced. b) half-life - The time it takes for the initial concentration of a reactant to decrease to half (50 %) of its original value. c) reaction intermediate - A substance that is produced in one step of a reaction mechanism and which is consumed in another step in the mechanism. Reactants, products, and catalysts are not reaction intermediates.
  2. (19.10) Write the reaction rate expressions for the following reactions in terms of the disappearance of the reactants and the appearance of products. a) 2 H 2 (g) + O 2 (g)  2 H 2 O(g) rate = - 1 / 2 ([H 2 ]/t) = - ([O 2 ]/t = 1 / 2 ([H 2 O]/t) b) 4 NH 3 (g) + 5 O 2 (g)  4 NO(g) + 6 H 2 O(g) rate = - 1 / 4 ([NH 3 ]/t) = - 1 / 5 ([O 2 ]/t) = 1 / 4 ([NO]/t) = 1 / 6 ([H 2 O]/t)
  3. Consider the reaction 2 N 2 O(g)  2 N 2 (g) + O 2 (g) a) Express the rate of reaction with respect to each of the reactants and products. rate = - 1 / 2 ([N 2 O]/t) = 1 / 2 ([N 2 ]/t) = [O 2 ]/t b) In the first 15.0 s of the reaction, 0.015 mol of O 2 is produced in a reaction vessel with a volume of 0.500 L. What is the average rate of the reaction over this time interval (including correct units)? The concentration of O 2 produced is [O 2 ] = 0.015 mol = 0.030 mol/L 0.500 L average rate = [O 2 ] =0.030 mol/L = 0.0020 mol/Ls t 15.0 s
  1. A reaction in which A, B, and C react to form products is zero order in A, second order in B, and first order in C. a) Write a rate law for the reaction. rate = k[B]^2 [C] b) What is the overall order of the reaction? Overall order = 0 + 2 + 1 = 3rd^ order c) By what factor does the reaction rate change if [A] is doubled, and the other reactants are held constant? No change d) By what factor does the reaction rate change if [B] is doubled, and the other reactants are held constant? Rate increases by 2^2 = 4 times faster. e) By what factor does the reaction rate change if [C] is doubled, and the other reactants are held constant? Rate increases by 2^1 = 2 times faster. f) By what factor does the reaction rate change if the concentrations of all three reactants are doubled? Rate increases by 2^02221 = 8 times faster.
  2. (19.30) The thermal decomposition of phosphine (PH 3 ) into phosphorus and molecular hydrogen is a first order reaction 4 PH 3 (g)  P 4 (g) + 2 H 2 (g) The half-life of the reaction is t1/2 = 35.0 s at T = 680. C. Find the following: a) The first order rate constant for the reaction. To calculate the rate constant, k , from the half-life of a first-order reaction, we use the relationship t1/2 = ln(2) k

The term k(0.0100 M)q^ appears in both the numerator and denominator, and so cancels, leaving (14.6 x 10-^5 /3.6 x 10-^5 ) = (0.0200/0.0100)p 4.06 = 2p^ p = 2 To find the reaction order for B we can compare trials 3 and 1 rate 3 = 7.0 x 10-^5 mol/Lmin = k[A 3 ]p[B 3 ]q^ = k(0.01000 M)p(0.0200 M)q rate 1 3.6 x 10-^5 mol/Lmin k[A 1 ]p[B 1 ]q^ k(0.0100 M)p(0.0100 M)q The term k(0.0100 M)p^ appears in both the numerator and denominator, and so cancels, leaving (7.0 x 10-^5 /3.6 x 10-^5 ) = (0.0200/0.0100)q 1.94 = 2q^ q = 1 Notice that we have picked p and q as the integer values that give the best agreement with the experimental results. Our rate law, therefore, is rate = k[A]^2 [B] (second order in A, first order in B, third order overall. b) The rate constant for the reaction (including correct units). If we solve our rate law for k, we get k = rate [A]^2 [B] Using the data from the first trial, we get k = (3.6 x 10-^5 mol/Lmin) = 36.0 L^2 /mol^2 min (0.0100 M)^2 (0.0100 M) Note that if we calculated k from one of the other trials we would get a slightly different value (36.5 L^2 /mol^2 min for trial 2, 35.0 L^2 /mol^2 min for trial 3). If this was a real experiment we would take the average of the three experimental values. On an exam, however, you can just pick one data set to calculate k. c) The initial rate of the reaction when [A] 0 = 0.01500 M, [B] 0 = 0.0300 M. rate = k[A]^2 [B] = (36.0 L^2 /mol^2 min)(0.0150 M)^2 (0.0300 M) = 20.2 x 10-^5 mol/Lmin

  1. The data below show the concentration of N 2 O 5 verses time for the reaction N 2 O 5 (g)  NO 3 (g) + NO 2 (g) time (s) [N 2 O 5 ] (M) time (s) [N 2 O 5 ] (M) 0.0 1.000 125.0 0. 25.0 0.822 150.0 0. 50.0 0.677 175.0 0. 75.0 0.557 200.0 0. 100.0 0. Determine the order of the reaction, the value for the rate constant (including correct units), and the half-life for the reaction. Predict the concentration of N 2 O 5 at 250.0 s. The general procedure used in these types of problems is as follows:
  2. plot ln (concentration) vs time. A straight line means a 1st^ order reaction.
  3. plot 1/(concentration) vs time. A straight line means a 2 nd^ order reaction.
  4. plot concentration vs time. A straight line means a 0th^ order reaction. If none of these gives a straight line, then the reaction order is different than 1st, 2 nd, or 0th^ order. (Note that there are more general methods that can be used to determine the experimental order, but they have not been discussed in the book and so will not be discussed here). The values needed to do the necessary plots are given in the table below. time [N 2 O 5 ] ln [N 2 O 5 ] 1/[N 2 O 5 ]
  1. 1.0 00 0.000 1.
  2. 0.822 - 0.196 1.
  3. 0.677 - 0.390 1.
  4. 0.557 - 0.585 1.
  5. 0.458 - 0.781 2.
  6. 0.377 - 0.976 2.
  7. 0.310 - 1.171 3.
  8. 0.255 - 1.366 3.
  9. 0.210 - 1.561 4.
  1. The following reaction was monitored as a function of time AB  A + B A plot of 1/[AB] vs time yields a straight line with slope m = 0.055 L/mol.s. a) What is the value of the rate constant, k, (including correct units) for this reaction at this temperature? The plot of 1/[AB] vs time gives a straight line. Therefore, the reaction is second order, and so k = 0.055 L/mols. b) Write the rate law for the reaction. rate = k [AB]^2 also [AB]t = [AB] 0 1 + kt [AB] 0 c) What is the half-life when the initial concentration is 0.55 mol/L? For a second order reaction t1/2 = 1/k[AB] 0. So t1/2 = 1. = 33.1 s (0.055 L/mols) (0.55 mol/L) d) If the initial concentration of AB is 0.250 mol/L, and the reaction initially contains no products, what are the concentrations of A and B after 75.0 s? [AB] = (0.250 mol/L) = 0.123 mol/L 1 + (0.055 L/mols) (75 s) (0.250 mol/L) Since no A or B were initially present in the system, the concentration of these species at t = 75. s is the same as the concentration of AB that has disappeared. Therefore [A] = [B] = 0.250 M - 0.123 M = 0.127 M
  1. The diagram below shows the energy of the reaction as the reaction progresses. Label each of the following in the diagram (reactants, products, activation energy, enthalpy of reaction). Also, is the reaction exothermic or endothermic? How do you know? The figure, with labels, is given below. Since energy decreases as we go from reactants to products, the reaction is exothermic (Hrxn < 0)
  2. (19.49) The rate constants are given for the first order reaction 2 N 2 O 5 (g)  2 N 2 O 4 (g) + O 2 (g) in the following table. Determine graphically the activation energy and pre-exponential factor for the reaction (including correct units). T (K) k (s-^1 ) T (K) k (s-^1 )
  1. 1.74 x 10-^5 328. 7.59 x 1 0 -^4
  2. 6.61 x 10-^5 338. 2.40 x 10-^3
  3. 2.51 x 10-^4 If we assume the Arrhenius equation applies, then we need to plot ln(k) vs 1/T. The data are given below T (K) 1/T (K-^1 ) k(s-^1 ) ln(k)
  4. 0.003356 1.74 x 10-^5 - 10.
  5. 0.003247 6.61 x 10-^5 - 9.
  6. 0.003145 2.51 x 10-^4 - 8. 328 0.003049 7.59 x 10-^4 - 7.
  7. 0.002959 2.40 x 10-^3 - 6. The data are plotted on the next page.

If we use the rate constant at T = 300. K, we get A = (2.6 x 10-^4 L/mols) exp[(9.28 x 10^4 J/mol)/(8.314 J/molK)(300 K)] = 3.8 x 10^12 L/mols

  1. The rate law for the reaction 2 NO(g) + Cl 2 (g)  2 NOCl(g) is given by the expression rate = k [NO][Cl 2 ]. a) What is the overall order of the reaction? Overall reaction order is 1 + 1 = 2nd^ order overall. b) One proposed mechanism for the above reaction is the following: step 1 NO(g) + Cl 2 (g)  NOCl 2 (g) (slow) step 2 NOCl 2 (g) + NO(g)  2 NOCl(g) (fast) Is the above mechanism consistent with the proposed rate law? Explain For the proposed mechanism the overall rate of reaction will be the rate of the slow step, and so rate = k 1 [NO][Cl 2 ] which is first order in NO and first order in Cl 2. This is the same as the experimental reaction orders, and so the mechanism is consistent with the experimental data. Note that this does not prove the mechanism is correct, as there could be other mechanisms that lead to a rate law with the same reaction orders.
  2. (19.63) The rate law for the reaction 2 H 2 (g) + 2 NO(g)  N 2 (g) + 2 H 2 O(g) is rate = k [H 2 ][NO]^2. Which of the following is a possible mechanism for the reaction (Hint: Find the rate law corresponding to each mechanism, and see whether or not it agrees with the experimental rate law.)

Mechanism I Step 1 H 2 (g) + NO(g)  H 2 O(g) + N(g) (slow) Step 2 N(g) + NO(g)  N 2 (g) + O(g) (fast) Step 3 O(g) + H 2 (g)  H 2 O(g) (fast) Mechanism II Step 1 H 2 (g) + 2 NO(g)  N 2 O(g) + H 2 O(g) (slow) Step 2 N 2 O(g) + H 2 (g)  N 2 (g) + H 2 O(g) (fast) Mechanism III Step 1 2 NO(g)  N 2 O 2 (g) (fast, reversible) Step 2 N 2 O 2 + H 2 (g)  N 2 O(g) + H 2 O(g) (slow) Step 3 N 2 O(g) + H 2 (g)  N 2 (g) + H 2 O(g) (fast) Mechanism I. In Mechanism I the slow step is bimolecular and the rate law would be: rate = k 1 [H 2 ][NO] Mechanism I can be discarded. Mechanism II. The rate-determining step in Mechanism II involves the simultan- eous collision of two NO molecules with one H 2 molecule. The rate law would be: rate = k 1 [H 2 ][NO]^2 Mechanism II is a possibility. Mechanism III. In Mechanism III the slow step is bimolecular and involves collision of a hydrogen molecule with a molecule of N 2 O 2. The rate would be: rate = k 2 [H 2 ][N 2 O 2 ] But [N 2 O 2 ] is a reaction intermediate. We may use the first step in the mechanism to find an expression for [N 2 O 2 ] k 1 [NO]^2 = k (^) - 1 [N 2 O 2 ] [N 2 O 2 ] = (k 1 /k (^) - 1 )[NO]^2