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KEY


                           GENERAL CHEMISTRY-II (1412)
                                    S.I. # 13

1. The 1st order rate constant for the decomposition of N2O5(g)
        N2O5 (g)  2 NO2 (g) + O2 (g)
at 70°C is 6.82x10-3s-1. Start with 0.0250 mol N2O5 (g) in a volume of 2.0 L.
a. How many mole of N2O5 will remain after 2.5 min.?
b. How many minutes will it take for the quantity of N2O5 to drop to 0.010 mol?
c. What is the half life of N2O5 at 70°C?
        ln [A] = -kt + ln [A]o
a. 2.5 min = 150s       [N2O5]o = (0.0250 mol/ 2 L) = 0.0125M
  ln [N2O5]150 = -(6.82x10-3s-1)(150s) + ln (0.0125) = -5.41
 [N2O5]150 = 4.494x10-3 = 4.5x10-3M; mol N2O5 = 4.494x10-3 x 2L = 9.0x10-3 mol
b. [N2O5]t = 0.010 mol/ 2L = 0.005M; [N2O5]o = 0.0125M
  ln (0.005) = -(6.83x10-3s-1)(t) + ln (0.0125 )
         t = -[ln (0.005) – ln (0.0125)] = 134.35s / 60s = 2.2 minutes
                6.82x10-3s-1
c. t1/2 = 0.693/k = 0.693/6.82x10-3s-1 = 101.6 or 102 seconds or 1.69 minutes

2. A reaction A + B  C obeys the following rate law: Rate = k[A] 2 [B].
        a. if [A] is doubled how will the rate change? Explain
        b. what are the reaction orders for A and B and the reaction overall?
        c. what are the units of k?
a. If [A] doubles, rate increases by a factor of 4, (k) is unchanged, rate is proportional to
[A]2 so 22 = 4, (k) only changes if Temperature changes.
b. Reaction order is 2nd order in [A], 1st order in [B], and 3rd order overall.
c. Units of k = M/s = M-2s-1
                M3
KEY


3. The decomposition of N2O5 in carbon tetrachloride proceeds as follows:
       2 N2O5  4 NO2 + O2 . The rate law is 1st order in N2O5. At 64°C the rate
constant is 4.82x10-3 s-1.
a. write the rate law for the reaction
b. what is the rate of the reaction when [N2O5] = 0.0240 M?
c. What happens to the rate when the concentration of N2O5 is doubled to 0.0480M?
       a. rate = k[N2O5] = 4.82x10-3 s-1     [N2O5]
       b. rate = 4.82x10-3 s-1 (0.0240M) = 1.16x10-4 M/s
       c. rate = 4.82x10-3 s-1 (0.0480M) = 2.31x10-4 M/s


4. 2 NO  2H2 (g) + 2H2O (g)
The rate law for this reaction is 1st order in H2 and 2nd order in NO. a. Write the
rate law. b. If the rate constant at 1000K is 6.0x104M-2s-1, what is the reaction rate
when [NO] = 0.025 M and [H2] = 0.015M? c. What is the reaction rate when the
concentrations are both 0.010M?
       a. rate = k[H2][NO]2
       b. rate = (6.0x104M-2s-1) (0.015M) (0.025M)2= 0.56 M/s
       c. rate = (6.0x104M-2s-1)(0.010M)(0.010M)2 = .06 M/s
5. What is the law of mass action? Give the formula and explain.
       The law of mass action expresses the relationship between the concentrations of
reactants and products at equilibrium for any action.
       Ex: Kc = [NOBr2] / [NO][Br2]


6. Write the equilibrium constant (Kc) for the following reactions:
       a. 2 O3 (g)  3 O2 (g)
       b. 2 NO (g) + Cl2 (g)   2 NOCl(g)
       c. Ag+ (aq) + 2 NH3 (aq)   Ag (NH3)2+ (aq)


       a. Kc = [O2]3 / [O3]2
       b. Kc = [NOCl]2 / [NO]2[Cl2]
       c. Kc = [Ag(NH3)2+] / [Ag+][NH3]2
KEY


7. Which of the following reactions lies to the right, favoring the formation of
products, and which lies to the left, favoring the reactants?
       a. 2NO (g) O2 (g)   2 NO2 (g) ; Kc = 5.0x1012
       b. 2 HBr (g)   H2 (g) + Br2 (g) ; Kc = 5.8x10-18


       a. Eq is to R (Kc >> 1) favors products
       b. Eq is to L (Kc << 1) favors reactants
chapter 15 #16

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SI #13 Key

  • 1. KEY GENERAL CHEMISTRY-II (1412) S.I. # 13 1. The 1st order rate constant for the decomposition of N2O5(g) N2O5 (g)  2 NO2 (g) + O2 (g) at 70°C is 6.82x10-3s-1. Start with 0.0250 mol N2O5 (g) in a volume of 2.0 L. a. How many mole of N2O5 will remain after 2.5 min.? b. How many minutes will it take for the quantity of N2O5 to drop to 0.010 mol? c. What is the half life of N2O5 at 70°C? ln [A] = -kt + ln [A]o a. 2.5 min = 150s [N2O5]o = (0.0250 mol/ 2 L) = 0.0125M ln [N2O5]150 = -(6.82x10-3s-1)(150s) + ln (0.0125) = -5.41 [N2O5]150 = 4.494x10-3 = 4.5x10-3M; mol N2O5 = 4.494x10-3 x 2L = 9.0x10-3 mol b. [N2O5]t = 0.010 mol/ 2L = 0.005M; [N2O5]o = 0.0125M ln (0.005) = -(6.83x10-3s-1)(t) + ln (0.0125 ) t = -[ln (0.005) – ln (0.0125)] = 134.35s / 60s = 2.2 minutes 6.82x10-3s-1 c. t1/2 = 0.693/k = 0.693/6.82x10-3s-1 = 101.6 or 102 seconds or 1.69 minutes 2. A reaction A + B  C obeys the following rate law: Rate = k[A] 2 [B]. a. if [A] is doubled how will the rate change? Explain b. what are the reaction orders for A and B and the reaction overall? c. what are the units of k? a. If [A] doubles, rate increases by a factor of 4, (k) is unchanged, rate is proportional to [A]2 so 22 = 4, (k) only changes if Temperature changes. b. Reaction order is 2nd order in [A], 1st order in [B], and 3rd order overall. c. Units of k = M/s = M-2s-1 M3
  • 2. KEY 3. The decomposition of N2O5 in carbon tetrachloride proceeds as follows: 2 N2O5  4 NO2 + O2 . The rate law is 1st order in N2O5. At 64°C the rate constant is 4.82x10-3 s-1. a. write the rate law for the reaction b. what is the rate of the reaction when [N2O5] = 0.0240 M? c. What happens to the rate when the concentration of N2O5 is doubled to 0.0480M? a. rate = k[N2O5] = 4.82x10-3 s-1 [N2O5] b. rate = 4.82x10-3 s-1 (0.0240M) = 1.16x10-4 M/s c. rate = 4.82x10-3 s-1 (0.0480M) = 2.31x10-4 M/s 4. 2 NO  2H2 (g) + 2H2O (g) The rate law for this reaction is 1st order in H2 and 2nd order in NO. a. Write the rate law. b. If the rate constant at 1000K is 6.0x104M-2s-1, what is the reaction rate when [NO] = 0.025 M and [H2] = 0.015M? c. What is the reaction rate when the concentrations are both 0.010M? a. rate = k[H2][NO]2 b. rate = (6.0x104M-2s-1) (0.015M) (0.025M)2= 0.56 M/s c. rate = (6.0x104M-2s-1)(0.010M)(0.010M)2 = .06 M/s 5. What is the law of mass action? Give the formula and explain. The law of mass action expresses the relationship between the concentrations of reactants and products at equilibrium for any action. Ex: Kc = [NOBr2] / [NO][Br2] 6. Write the equilibrium constant (Kc) for the following reactions: a. 2 O3 (g)  3 O2 (g) b. 2 NO (g) + Cl2 (g)   2 NOCl(g) c. Ag+ (aq) + 2 NH3 (aq)   Ag (NH3)2+ (aq) a. Kc = [O2]3 / [O3]2 b. Kc = [NOCl]2 / [NO]2[Cl2] c. Kc = [Ag(NH3)2+] / [Ag+][NH3]2
  • 3. KEY 7. Which of the following reactions lies to the right, favoring the formation of products, and which lies to the left, favoring the reactants? a. 2NO (g) O2 (g)   2 NO2 (g) ; Kc = 5.0x1012 b. 2 HBr (g)   H2 (g) + Br2 (g) ; Kc = 5.8x10-18 a. Eq is to R (Kc >> 1) favors products b. Eq is to L (Kc << 1) favors reactants chapter 15 #16