Gaseous phase decomposition of N2O is given below. Step I : N2O(g) →N2(g) + O(g) Step II : N2O(g) + O(g)→ N2(g) + O2(g) (a) Write a chemical equation for overall reaction. (b) Identify the reaction intermediate. (c) What is the molecularity of each of the elementary reaction?
Answers
Answer:
Reaction Mechanisms
17. 2 NO2 (g) + F2 (g) 2 FNO2 (g) rate = k [NO2][ F2]
a. Describe the collisions that would need to occur for this reaction to occur in a single step. How many
molecules must collide? Explain why it is unlikely.
Three molecules would need collide simultaneously. Not likely.
b. What would the rate law be if the reaction occurred in one step? How does the experimentally
determined rate law eliminates this possibility.
rate = k [NO2]
2
[ F2] – in elementary steps the rate law is determined by the stoichiometry of the
reaction as opposed to the overall rate law that MUST be determined experimentally.
c. The proposed mechanism includes two steps. Identify the rate limiting step. Justify your reasoning.
Elementary Step 1: NO2 (g) + F2 (g) FNO2 (g) + F (g) (slow , higher Ea smaller k1)
Elementary Step 2: NO2 (g) + F (g) FNO2 (g) (fast , lower Ea larger k2)
The rate limiting step is the slow step, since the reaction can only go as fast as the slowest step.
b. Explain how the proposed mechanism is consistent with the experimental rate law.
Elementary Step 1: NO2 (g) + F2 (g) FNO2 (g) + F (g) rate = k [NO2][ F2]
(slow , Ea smaller - k1) NO2 must collide with F2 in a single collision
Elementary Step 2: NO2 (g) + F (g) FNO2 (g) rate = k [NO2][ F]
(fast , Ea larger - k2) NO2 must collide with F in a single collision
The steps add up to the overall reaction since the F cancels out as an intermediate.
The rate limiting step shows a rate law that is consistent with the experimental determined rate law
e. Explain why the rate limiting step has a greater activation energy, Ea and lower rate constant, k.
Higher activation energy – will require more energy in order for a collision to be effective thus a smaller
fraction of collisions will be effective
18. The Raschig reaction produces the industrially important reducing agent hydrazine (N2H4), from NH3 and
OClin a basic aqueous solution
Step 1: Fast NH3 (aq) + OCl-
(aq) NH2Cl (aq) + OH-
(aq)
Step 2: Slow NH2Cl (aq) + NH3 (aq) N2H5
+
(aq) + Cl-
(aq)
Step 3: Fast N2H5
+
(aq) + OH-
(aq) N2H4 (aq) + H2O (l)
a. What is the overall stoichiometric equation? Hint: combine the reactions and cross out intermediates.
2 NH3 (aq) + OCl-
(aq) N2H4 (aq) + H2O (l)
b. Which step is the rate-limiting step? How did you pick this step from among the three?
Step 2 is the rate limiting step since it is the slow step
and reaction can only go as fast as the slowest step
c. Write the rate equation for the rate-limiting step.
Rate = k [NH2Cl][ NH3] for step 2
d. What reaction intermediates is included in the rate law for the slow step?
NH2Cl created in step 1 and reacts in step 2
e. The overall rate law may not include an intermediate. Adjust the rate law for the slow step to replace
the intermediate. This is the rate law for the overall reaction.
Rate law for step 1 = Rate = k [ NH3][OCl-] this must be substituted into the rate law for step 2 to get the
overall NOT in terms of an intermediate
Rate = k [NH2Cl][ NH3] for step 2
Rate = k [ NH3][OCl-] for step 1 that creates intermediate
Put rate law for step 1 into rate law for step 2 putting in place of the NH2Cl intermediate
Rate = k [ NH3][OCl-][ NH3]