CHEMISTRY 1AA3
TUTORIAL PROBLEM SET 6 Week of FEBRUARY 25, 2002
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Answers NB QUESTION 3, 5, 6 HAD MINOR ERRORS LOOK FOR RED
CORRECTED Mar. 12, 2002
1.
The experimental rate law for the reaction:
2 ICl (g) + H2 (g) → 2HCl (g) + I2 (g)
is Rate = k [ICl][H2]
Suggest a mechanism which is consistent with these observations. While you can use as many
steps as you like, there is a mechanism with two steps.
One acceptable answer:
ICl (g) + H2 (g) → HCl (g) + HI (g)
ICl (g) + HI (g) → HCl (g) + I2 (g)
slow (RDS)
2. A reaction that contributes to smog formation is the combination of oxygen with NO from
automobile exhaust.:
2 NO (g) + O2 (g) → 2NO2 (g)
Show that the observed rate law:
rate = k [NO]2 [O2]
is consistent with the following mechanism.
1 NO (g) + O2 (g)
k1
NO3 (g)
k-1
2
NO3 (g) + NO (g)
k2
2NO2 (g)
THE ANSWER
Assume the overall rate = ½ d[NO2]/dt = k2 [NO2][NO3] (i.e. second step is RDS)
However, we don’t know the concentration of NO3. Assuming the steady state approximation for
the equilibrium is valid, that is, the concentration of NO3 is invariant with time, we can solve for
[NO3]
K = [products]/[reactants]
for equation 1, K = [NO3]/[NO][O2]
∴[NO3] = K [NO][O2]
Substituting this into the rate equation, we get
rate = k2 [NO2][NO3] = k2 [NO] K [NO][O2] = k2K [NO]2[O2] = kobs [NO]2[O2]
3. The conversion of gaseous cis-2-butene to trans-2-butene occurs in one step with ∆H° = -4.4
kJ/mol and Ea = 218 kJ/mol at 25 °C.
i. Draw a potential energy profile for this reaction
ii) Calculate the Ea for the reverse reaction.
iii) Assuming the preexponential factor A is the same for this reaction in both directions and that
the reaction is first order in trans-2-butene, how much slower (or faster) will the reverse reaction
take place at 400°C?
ANSWER
i)
Energy
A
218
cis
222.4
4.4
trans
Reaction Coordinate
ii) =218 + 4.4 = 222.4 kJ/mol
iii) kforward = A e-218/R673
kreverse = A e-222.4/R673
The ratio of rate constants tells how much faster one will go than the other
kforward / kreverse = A e-218/R673 / A e-222.4/R673
k = Ae-Ea/RT
kforward / kreverse = e-218/R673+222.4/R673
= 4.4 kJ/mol /673 K x 8.314 J /mol K
e
e
= 4400 J/mol /673 K x 8.314 J /mol K
= 2.19
The forward reaction goes 2.19 times faster than the reverse reaction
4. The activation energy for the decomposition of ammonia (NH3) is 335 kJ/mol for the
uncatalyzed reaction and 165 kJ/mol when the reaction is catalyzed by powdered tungsten.
The enthalpy change for the reaction is 46 kJ/mol. Calculate the molar activation energy for
the formation of NH3 from its elements when the reaction is:
i. catalyzed
ii. uncatalyzed
NH3 (g) → ½ N2 (g) + 3/2 H2 (g)
ANSWER
The reaction is endothermic. As shown on the graph, the activation energy for the reverse
reaction is 46 kcal less than the forward reaction. 335-46 = 289 kJ/mol. Similarly the catalyzed
reaction will have an activation energy of 119 kJ/mol.
335-46 = 289 kJ/mol
Energy
A
335
N2/H2
46
NH3
Reaction Coordinate
5.
The half life of a first order reaction (to be named later) is 20.0 s at 25 °C and 1.0 s at 50
°C. Calculate the activation energy and preexponential factor A for this reaction.
ANSWER
k = ln 2/t½ k at 25 °C k = 0.035; at 50 °C k = 0.69
k = Ae-Ea/RT
Plot of versus ln k versus 1/T gives slope -Ea/R and intercept = lnA
Ea = 96 kJ/mol, A = 2.2 x 1015 s-1
6. Consider the following reaction mechanism
H2O2 → H2O + O
O + CF2Cl2 → ClO + CF2Cl
ClO + O3 → Cl + 2O2
Cl + CF2Cl → CF2Cl2
NET REACTION
H2O2 + O + CF2Cl2 + ClO + O3 + Cl + CF2Cl → H2O + O + ClO + CF2Cl + Cl + 2O2 +
CF2Cl2
reduces to
H2O2 + O3 → H2O + 2O2
i) What is the molecularity of each step?
ii) Write the overall equation for the reaction.
iii) Identify the reaction intermediate(s).
iv) What is the role of CF2Cl2?
v) Write the overall rate law assuming the first step is the rate determining step.
vi) Write the overall rate law assuming the second step is the rate determining step and the
first step is actually a rapid equilibrium (steady state approximation).
ANSWER
i) Molecularity of an elementary step is like order of an overall process. The steps are
unimolecular, bimolecular, bimolecular, bimolecular, respectively.
ii) see reduced equation above
iii) the reaction intermediates are in italics
iv) it is a catalyst. It is neither consumed nor created.
v) Rate = k [H2O2]
vi) Rate = k2 K1 [H2O2][CF2Cl2] = kobs[H2O2][CF2Cl2]/[H2O]