Exam 1B, Spring 2018
Exam 1B, Spring 20181. Give the chemical formula for the nitrite ion.
NO2–
2. Give the name for CrO42–.
chromate ion
3. What are the units of the rate constant of a second order reaction?
M–1s–1
4. The rate law for the reaction shown below is first order in each reactant with k = 3.0×10–4 M–1s–1 at 25 °C. Calculate the rate of reaction at 25 °C if [NH4+] = 0.50 M and [NO2–] = 0.050 M.
NH4+(aq) + NO2–(aq) → N2(g) + 2 H2O(l)
Rate = k[NH4+][NO2–] = (3.0×10–4)(0.50)(0.050) = 7.5×10–6 M·s–1
5. Consider the first order reaction: PH3(g) → P4(g) + H2(g).
a. Balance the reaction.
b. The half-life is 35.0 s at 680 °C. Find the rate constant, including units.
a. 4 PH3(g) → P4(g) + 6 H2(g)
b. For first order reactions k = 0.693/t½ = 0.693/35.0 = 0.0198 s–1
6. The mechanism shown below is proposed for the reaction Tl+(aq) + Ce4+(aq) → Tl3+(aq) + Ce3+(aq) that occurs only in the presence of Mn2+(aq) with a rate law of Rate = k[Ce4+][Mn2+].
a. Is the mechanism plausible? Why or why not?
b. What is the role of the Mn2+?
Ce4+(aq) + Mn2+(aq) → Ce3+(aq) + Mn3+(aq)
Ce4+(aq) + Mn3+(aq) → Ce3+(aq) + Mn4+(aq)
Tl+(aq) + Mn4+(aq) → Tl3+(aq) + Mn2+(aq)
a. First, the reaction must be balanced:
Tl+(aq) + 2 Ce4+(aq) → Tl3+(aq) + 2 Ce3+(aq) (this is required to balance charge)
Summing the reactions in the proposed mechanism and eliminating common species gives:
Ce4+(aq) + Mn2+(aq) + Ce4+(aq) + Mn3+(aq) + Tl+(aq) + Mn4+(aq) → Ce3+(aq) + Mn3+(aq) + Ce3+(aq) + Mn4+(aq) + Tl3+(aq) + Mn2+(aq)
Tl+(aq) + 2 Ce4+(aq) → Tl3+(aq) + 2 Ce3+(aq)
Which matches the observed reaction, so the stoichiometry requirement is met.
All of the steps are bimolecular so identifying the rate-determining step is not obvious. If step 1 is rate-determining, then the rate law will match the observed rate law, which is plausible.
Thus, the mechanism is plausible.
b. Mn2+ is a catalyst.
7. Find the rate law and average rate constant (including units) for the reaction run at 700 °C: 2 H2(g) + 2 NO(g) → 2 H2O(g) + N2(g)
Experiment [H2] (M) [NO] (M) Initial Rate (M·s–1)
1 0.020 0.025 4.8×10–6
2 0.010 0.025 2.4×10–6
3 0.020 0.0125 1.2×10–6
The rate law is written as Rate = k[H2]m[NO]n
To find the orders of reaction use the Method of Initial Rates.
To find m use experiments 1 and 2: when the H2 concentration doubles the initial rate also doubles, which means that m = 1.
To find n use experiments 1 and 3: when the NO concentration doubles the initial rate quadruples, which means that n = 2.
Thus, the rate law is Rate = k[H2][NO]2
To find the average rate constant, evaluate k for each experiment using the concentrations and experimental rates:
Experiment 1: k = Rate1/[H2]1[NO]12 = 4.8×10–6/(0.020)(0.025)2 = 0.38 M–2s–1
Experiment 2: k = Rate2/[H2]2[NO]22 = 2.4×10–6/(0.010)(0.025)2 = 0.38 M–2s–1
Experiment 3: k = Rate3/[H2]3[NO]32 = 1.2×10–6/(0.020)(0.0125)2 = 0.38 M–2s–1
Conveniently, all of the experiments give the same value for the rate constant so the average is k = 0.38 M–2s–1
8. The rate constant for the reaction 2 N2O5(g) → 2 N2O2(g) + O2(g) was measured at several temperatures, as given below. Plot (using best practices for graphing) the data so that you could find the activation energy. A correct plot will have a slope of –12143; find the activation energy in units of kJ/mol. What is the order of reaction?
T °C k (s–1)
0 7.87×103
25 3.46×105
45 4.98×106
65 4.87×107
To obtain the activation energy a plot of T–1 (in units of K–1) vs lnk is used:
T °C k (s–1) T (K) T–1 (K–1) lnk (lns–1)
0 7.87×103 273 0.00366 8.971
25 3.46×105 298 0.00336 12.754
45 4.98×106 318 0.00314 15.421
65 4.87×107 338 0.00296 17.701
Plot:
The slope is used to find the activation energy: Ea = –slope×R = –(–12143)(8.314) = 101000 J/mol = 101 kJ/mol.
Based on the units of the rate constant the reaction is first order.
