1. The thermal decomposition of acetaldeh mechanism: yde. CHCHO CH, + CO, has th
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1. The thermal decomposition of acetaldeh mechanism: yde. CHCHO CH, + CO, has the following proposed hu (1) ki I got question #1 (3) k (4) ka (5) ks ng, termination, etc. What could hu represent? ust nee (a) Identify eaeh step as initiation, branchi (b) Write an equation for d CH Jdt. Apply the steady state approximation to write an equation for [CH*). You might need to use the SSA result for CHsCO to solve for [CH ).(Yes, it's a wee bit messy, but overall, lovely. Just lovely.). 2. Consider the following mechanism for the decomposition of NO ki NO:+NO -1 k2 NO. + NO NO2 + NO + O2 ky NO + N2O3 3NO2 (a) Identify the reactants, the products and the intermediates. (b) Write the rate equations for the consumption of N Os and the production of NO (c) Write the rate equations for any intermediates. Use the steady state approximation to derive expressions for the steady state concentrations of the intermediates (d) Substitute the results of (c) into your equations from (b), and simplify. at is the order of the reaction? What relationship exists between the rate of loss of N Os and the rate of appearance of NO,? Is this consistent with the stoichiometry of the overall reaction? key mechanism as a freshman, but this mechanism does not explain the stability of the transition state. The "induced fit model" helps account for this stability. Look up and explain the mechanistic steps of the induced fit model. Discuss the energy and relative kinetics of the steps. Give an example of an enzyme that follows this mechanism. Cite your sources 3. Enzymes are biological catalysts of amazing specificity. You probably learned about the "lock andExplanation / Answer
a) N2O5 is the reactant. The reaction produces NO2 and O2; hence, these two are the products. The reaction gives NO3 and NO as intermediates; these intermediates are produced and consumed during the course of the reaction.
b) N2O5 is consumed in steps 1 and 3; however, step 1, being a reversible reaction re-produces N2O5. Thus, the rate of consumption (note that the term used is consumption; hence any production must be denoted with a negative sign) is
R1 = k1[N2O5] + k3[NO][N2O5] – k-1[NO2][NO3] ……..(1)
NO2 is produced in steps 1 and 3 and consumed in step 1. Note that step 2 consumes and produces NO2; therefore, NO2 most likely acts as a catalyst in this step and does not appear in the rate equation. Therefore, the rate of production of NO2 can be written as
R2 = k1[N2O5] + k3[NO][N2O5] - k-1[NO2][NO3] …….(2)
c) NO3 and NO are the intermediates. NO3 is produced in step 1 and consumed in steps 1 and 2. The rates of production and consumption are given as
R3 = k1[N2O5] (rate of production) ……(3)
R4 = k-1[NO2][NO3] + k2[NO3] (rate of consumption) ……(4)
According to the steady state approximation
R3 = R4
Therefore,
k1[N2O5] = k-1[NO2][NO3] + k2[NO3]
=====> [NO3] = k1[N2O5]/(k-1[NO2] + k2) ……..(5)
NO is produced in step 2 and consumed in step 3. The rates of production and consumption are
R5 = k2[NO3] (rate of production)……(6)
R6 = k3[NO][N2O5] ……(7)
Again, as per the steady state approximation,
R5 = R6
=====> k2[NO3] = k3[NO][N2O5]
=====> [NO] = k2[NO3]/(k3[N2O5]) = k2/(k3[N2O5])* k1[N2O5]/(k-1[NO2] + k2) = k1k2/k3(k-1[NO2] + k2) …….(8)
d) Use the expressions for [NO3] and [NO] from above into R1 and R2.
R1 = k1[N2O5] + k3[NO][N2O5] – k-1[NO2][NO3] = k1[N2O5] + k3* k1k2/k3(k-1[NO2] + k2)*[N2O5] – k-1[NO2]* k1[N2O5]/(k-1[NO2] + k2)
= k1[N2O5]*{1 + k2/(k-1[NO2] + k2) – k-1[NO2]/ (k-1[NO2] + k2)}
= k1[N2O5]*{k-1[NO2] + k2 + k2 – k-1[NO2]}/ (k-1[NO2] + k2)
= 2k1k2[N2O5]/ (k-1[NO2] + k2)
We have seen from above that R1 = R2 and hence, the rate of production of NO2 is
Rate of production = 2k1k2[N2O5]/ (k-1[NO2] + k2) (ans)
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