The reaction profiles of reaction 1 and reaction 2 are shown in the figure. Answ
ID: 959976 • Letter: T
Question
The reaction profiles of reaction 1 and reaction 2 are shown in the figure. Answer the following questions. What is E_a for reaction 1 and reaction 2. respectively? Which reaction is faster? Which reaction is endothermic? What is E_a for the reverse reaction 2? Which reverse reaction is faster? Based on the right reaction profile of reaction A rightarrow C, How many intermediates are formed? Which step is the limiting step? Is this reaction endothermic or exothermic? How many transition states (T.S.) are there? Label T.S. on the profile. What is the molecularity of each of the following elementary reactions? Write the rate law for each.Explanation / Answer
a) For reaction 1 Ea = 6.8 kj/mol and for reaction 2 Ea = (8.4 - 4.4) = 4 kj/mol
b) Reaction 1 is faster since it require less activation energy.
c) Reaction 2 is endo thermic since energy is consumed by reactants
d) Ea for reverse of reaction1 = (0.5 +6.8) = 7.3 kj/mol
e) Ea for reverse of reaction2 = 8.4 kj/mol
f) Reverse of reaction 1 is faster since it require less activation energy.
7.
a) B ------> C is limiting step because the step is slow step from the graph
b) B -----> C
c) Given reaction is endothermic because product energy is higher than the reactants
d) Two transition states one is between A and B and another one is between B and C
8.
A + B ----> C Rate law = k [A]^1[B]^1 so molecularity 1+1 = 2
2 SO3 (g) ----> 2 SO2 (g) + O2 (g) rate law = k [SO3]^2 so molecularity = 2
2 NO (g) ----> N2O2 (g) rate law = k [NO]^2 so molecularity = 2
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