Atkin's Chapter 25 Homework

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All data not from the question came from the appendices.
25E3b The dimensions of k are
dim of R / { (dim of [A])×(dim of [B])2 } 
    = { amount × length-3 × time-1 } / { amount×length-3 }3
    = length6 × amount-2 × time-1
In mol, L, s units, the units of k are L2 mol-2 s-1
  1. R = - d[A]/dt = k[A][B]2 so d[A]/dt = - k[A][B]2
  2. R = d[C]/dt so d[C]/dt = k[A][B]2
25E5b The rate law is
R = kpa = kp0a(1-f)a
where a is the reaction order and f the fraction reacted (so that 1-f is the fraction remaining). Thus,
R1/R2 = kp0a(1-f1)a / [ kp0a(1-f2)a ] = [(1-f1) / (1-f2)]a
and
a = ln(v1 / v2) / ln[(1-f1) / (1-f2)]
   = ln(9.71 / 7.67) / ln[(1-0.100) / (1-0.200)] = 2.00
25E8b From Table 25.3, we see that for
A + 2 B arrow right P
the integrated rate law is
kt = { 1 / ( [B]0 - 2[A]0 ) } ×
    ln[ { [A]0×([B]0 - 2x) } / { ([A]0 - x)×[B]0 } ]
  1. Substituting the data after solving for k,

    2. k = { 1 / [(3.6×103 s)×(0.080 - 2×0.075)×(mol/L)] } ×
        ln{ [0.075×(0.080 - 0.060)] / [(0.075 - 0.030)×0.080] }
       = 3.47×10-3 L mol-1 s-1
  2. The half-life in terms of A is

    3. t½ = { 1 / [k([B]0 - 2[A]0)] } ×
          ln ( { [A]0×([B]0 - 2×½[A]0) } / { ½[A]0[B]0 } )
    which reduces to
    t½ = { 1 / [ k([B]0 - 2[A]0) ] } × ln(2 - 2[A]0/[B]0)
        = { 1 / [(3.47×10-3 L/mol s-1)×(-0.070 mol/L)] } ×
              ln[ 2 - (0.150/0.080)]
        = 8561 s = 2.4 hours

    The half-life in terms of B is
    t½ = { 1 / [k([B]0 - 2[A]0)] } ×
            ln({ [A]0([B]0 - ½[B]0) } / { ([A]0 - ¼[B]0)×[B]0 })
    which reduces to
    t½ = { 1 / [k([B]0 - 2[A]0)] }×ln(½[A]0 / {[A]0 - ¼[B]0})     = { 1 / [(3.47×10-3 L/mol s-1)×(-0.070 mol/L)] } ×
            ln(½[0.075] / [0.075 - ¼(0.080)] }
        = 1576 s = 0.44 hour

25E13b The rate constant is given by
k = A e-Ea/RT
so at 50°C, it is
1.70×10-2 L/mol s-1 = A e-Ea / {(8.3145 J/mol K-1)×([24+273] K)}
and at 50°C, it is
2.01×10-2 L/mol s-1 = A e-Ea / {(8.3145 J/mol K-1)×([37+273] K)}
Dividing the two rate constants yields
(1.70×10-2)/(2.01×10-2) =
    e[-Ea/8.3145 J/mol K-1]×[(1/297 K) - (1/310 K)] }
so
ln(1.70/2.01) = [-Ea/8.3145 J/mol K-1]×[(1/297 K) - (1/310 K)]
and
Ea = - (8.3145 J/mol K-1) × [(1/297 K) - (1/310 K)]-1 ×
        ln(1.70/2.01)
    = 9.9×103 J/mol = 9.9 kJ/mol

With the activation energy in hand, the prefactor can be computed from either rate constant value
A = k e+Ea/RT
    = (1.70×10-2 L/mol s-1) e9.9×103 J/mol / (8.3145 J/mol K-1 × 297 K)
    = 0.94 L/mol s-1

25E17b The effective rate constant is related to the individual steps by
1/k = [ka'/(kakb)] + 1/[kap]
so
(1/k1) - (1/k2) = (1/ka)[(1/p1) - (1/p2)
ka = [(1/p1) - (1/p2)]×[(1/k1) - (1/k2)]-1
    = [(1/1.09×103 Pa) - (1/25 Pa)] ×
        [(1/1.7×10-3 s-1) - (1/2.2×10-4 s-1)]-1
    = 9.9×10-6 s-1 Pa-1
25P4 An examination of the data shows that the half-life remains constant at about 2 minutes. Therefore, the reaction is first-order. This can be confirmed by fitting any two pairs of data to the integrated first-order rate law, solving for k from each pair, and checking to see that they are the same to within experimental error.

ln([A]/[A]0) = - kt (10b, A=N2O5)
Solving for k,
k = ln([A]0/[A]) / t
at t=1.00 min, [A] = 0.705 mol/L
k = ln(1.000/0.705) / 1.00 min = 0.350 min-1 = 5.83×10-3 s-1
at t=3.00 min, [A] = 0.399 mol/L
k = ln(1.000/0.349) / 3.00 min = 0.351 min-1 = 5.85×10-3 s-1

Values of k may be determined in a similar manner at all other times. The average value of k obtained is 5.48×10-3 s-1. The constancy of k, which varies only between 5.83 and 5.85×10-3 s-1, confirms that the reaction is first-order. A linear regression of ln[A] against t yields the same result.

25P22 kt = [ 1 / (n-1) ]×[ (1/[A]n-1) - (1/[A]0n-1) ] (Ex. 25.15a, n<>1)
at t=t½, kt½ = [1/(n-1)]×[(2/[A]0)n-1 - (1/[A]0)n-1]
at t=t¾, kt¾ = [1/(n-1)]×[(4/3[A]0)n-1 - (1/[A]0)n-1]
Hence
t½/t¾ = [ 2n-1 - 1 ] / [ (4/3)n-1 - 1 ]
25P31
  1. The rate of the reaction is
    R = k [CH4][OH]
       = 1.3×109 L/mol s-1 e-14.1×103 J/mol/(8.3145 J/mol K-1×263 K)
            × (4×1021 L) × (365×24×3600 s)
       = 1.1×10-16 mol/L s-1

  2. The mass is the amount consumed (in moles) times the molar mass; the amount consumed is the rate of consumption times the volume of the "reaction vessel" times the time.
    m = MvVt
       = (0.01604 kg/mol)(1.1×10-16 mol/L s-1)(4×1021 L) ×
            (365××24×3600 s)
       = 2.2×1011 kg or 220 Tg
Last modified 24 November 1999