More practice problems for exam 4

32) Fill in the table using 0.1 M for all initial conditions and 1 s-1 or 1 M-1s-1 or 1 M-2s-1 as appropriate for all rate constants

Reaction

Differential rate equations

Initial rates (M/s)

Final concentrations

A --> B

dA/dt = -k[A]

dB/dt = k[A]

dA/dt = -0.1 M/s

dB/dt = 0.1 M/s

A =0 M

B = 0.2 M

A + B -- > C

dA/dt = -k[A][B]

dB/dt = -k[A][B]

dC/dt = k[A][B]

dA/dt = -0.01 M/s

dB/dt = -0.01 M/s

dC/dt = 0.01 M/s

A = 0 M

B = 0 M

C = 0.2 M

A <--> B

dA/dt = -kf[A] + kb[B]

dB/dt = kf[A] - kb[B]

dA/dt = 0 M/s

dB/dt = 0 M/s

A = 0.1 M

B = 0.1 M

A+B <--> C

dA/dt = -kf[A][B] + kb[C]

dB/dt = -kf[A][B] + kb[C]

dC/dt = kf[A][B] - kb[C]

dA/dt = 0.09 M/s

dB/dt = 0.09 M/s

dC/dt = -0.09 M/s

A = 0.1708 M

B = 0.1708 M

C = 0.0292 M

A --> B -->C

dA/dt = -k1[A]

dB/dt = k1[A] - k2[B]

dC/dt = k2[B]

dA/dt = -0.1 M/s

dB/dt = 0 M/s

dC/dt = 0.1 M/s

A = 0 M

B = 0 M

C = 0.3 M

A <--> B+C --> D

dA/dt = -kf1[A] + kb1[B][C]

dB/dt = kf1[A]- kb1[B][C] -k2[B][C]

dC/dt = kf1[A]- kb1[B][C] -k2[B][C]

dD/dt = k2[B][C]

dA/dt = -0.09 M/s

dB/dt = 0.08 M/s

dC/dt = 0.08 M/s

dD/dt = 0.01 M/s

A = 0 M

B = 0 M

C = 0 M

D = 0.3 M

2A +B <-->C-->D

dA/dt = 2(-kf1[A]2[B] + kb1[C])

dB/dt = -kf1[A]2[B] + kb1[C]

dC/dt = kf1[A]2[B] - kb1[C] - k2[C]

dD/dt = kf2[C]

dA/dt = 0.198M/s

dB/dt = 0.099 M/s

dC/dt =-0.199M/s

dD/dt = 0.1 M/s

A = 0 M

B = 0.05 M

C = 0 M

D = 0.25 M

A+B<-->C<--> D

dA/dt = -kf1[A][B] + kb1[C]

dB/dt = -kf1[A][B] + kb1[C]

dC/dt =kf1[A][B]-kb1[C]-kf2[C]+kb2[D]

dD/dt = kf2[C] - kb2[D]

dA/dt = 0.09 M/s

dB/dt = 0.09 M/s

dC/dt = -0.09 M/s

dD/dt = 0 M/s

A = 0.2109 M

B = 0.2109 M

C = 0.04451 M

D = 0.04451 M
  1. Consider the simple first order reaction, A --> B. If you start with 1 M A and 0 M B and it takes 10 seconds to go to 0.5 M A and 0.5 M B, what is the rate constant for this reaction?

    2. A = A0 exp(-k t) so 0.5M = (1 M) exp(-k 10s) or k = -ln(0.5) /10 s = 0.0693 s-1

     

  2. Consider the reaction A<-->B --> C. Let the first reaction have a forward rate constant of 1 s-1 and a reverse rate constant of 50 s-1 and let the second reaction have a rate constant of 5 s-1. If I initially add 1 mole of A to one liter of solvent (no B or C added), how long will it take before the concentration of C is 0.5 M using a) the equilibrium approximation and b) the steady state approximation?

    3. The first thing to recognize is that the equilibrium constant strongly favors the reactant A. Thus, there is never very much B present in the reaction. We can also see that

    dC/dt = 5 s-1 [B]. At equilibrium [B]/[A] = 1/50 so [B] = [A]/50 and therefore dC/dt =5 s-1 [A]/50 = 0.1 (s-1) [A] . Now, since there is little [B] present at any time,

    dA/dt ~ -dC/dt = -0.1(s-1) [A] or dA/[A] = -0.1 (s-1) dt. This can be integrated to give

    lnA = -0.1 (s-1) t + const., were const. is the constant of integration. As shown in the notes, this can be rearranged and evaluated to give A(t) = A0exp(-0.1s-1 t), where A0 is the initial value of A or 1M in this case since there is little B present at any time. Thus, to find how long it takes to get to the point where half the reaction is over (we started with 1 M A),

    0.5M = 1Mexp(-0.1 s-1 t) or

    t = ln(0.5) / (-0.1 s-1) = 6.935 seconds.

    The steady-state approximation is similar except [B]/[A] = kf1/(k2+kb1) = 1/55. Everything else is the same. You end up with t = ln(0.5)/(-0.0909) = 7.63 seconds. The steady state approximation is probably more accurate in this case since it includes the effect of considering the forward reaction from B to C.

     

     

  3. At 298 K the first order rate constant for the conversion of A to B was measured to be measured to be 10 s-1. At 330 K, the first order rate constant was measured to be 20 s-1. What is the activation energy (in joules per mole) for this reaction?

    4. At the two temperatures we can write

    Now just solve the two equations simultaneously for EA giving 17.71 kJ/mole.

     

  4. For the reaction A + B<--> C <--> D, the forward rate constant for the first reaction is 10 M-1s-1, the reverse rate constant for the first reaction is 1 s-1, the forward rate constant for the second reaction is 5 s-1 and the reverse rate constant is 0.1 s-1 at 298 K. What is the standard reaction free energy for the overall reaction A + B <--> D?

For the first reaction, the equilibrium constant is Keq1 = 10/1 = 10. The second equilibrium constant is Keq2 = 5/.1 = 50. The equilibrium constant for the overall reaction is just the product of these two since

Given the equilibrium constant for the overall reaction, we can calculate the standard reaction free energy as -RTlnKeq = -(8.314)(298)ln(500) = 15.40 kJ/mole.