When inputting these rate constants below, use "NO" for NO, "02" for O2, "N202" for N₂O2, "N204" for N2O4, "k_1" for k₁, "k_1rev" for k-1, "k_2" for k2, "k_2rev" for k-2, and "k_3" for ką. Without making any approximations, what is the rate of production of NO2?
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PLEASE COMPLETE QUESTION ANSWER THAN RATING HEPLFUL
Step by step
Solved in 4 steps
- The following mechanism is proposed: C6H5COOH ⇌ C6H5CO2H2+ (fast) C6H5CO2H2+ ⇌ C6H5CO+ + H2O (fast) C6H5CO+ + HX + H2O ⟶ C6H6NH2 + N2 + CO2 (slow) The overall rate constant for this reaction is 0.200x 10-3L mol-1 s-1. Is this value consistent with rate law r = kobs*[C6H5COOH]*[HX] ?Determine the molecularity of the rate limiting step in this mechanism: S1 (fast): C6H6 + Br2 + FeBr3 --> C6H5Br+ + FeBr4 - S2 (slow): C6H5Br+ + FeBr4 - --> C6H5Br + HBr + FeBr3 a.) unimolecular b.) bimolecular c.) trimolecular d.) can't be determinedSuppose the formation of nitryl fluoride proceeds by the following mechanism: step elementary reaction rate constant 1 NO2g+F2g)->NO2F(g)+F(g) k1 2 F(g)+NO2(g)->NO2F(g) k2 Suppose also k1 ≪ k2. That is, the first step is much slower than the second.
- . The proposed reaction mechanism is as follows:i. BrO3- (aq) + H+ (aq) à HBrO3 (aq) [Fast]ii. HBrO3 (aq) + H+ (aq) à H2BrO3+ (aq) [Medium]iii. H2BrO3+ (aq) + Br- (aq) à Br2O2 (aq) + H2O (ℓ) [Slow]iv. Br2O2 (aq) + 4H+(aq) + 4Br-(aq) à 3Br2 (ℓ) + H2O (ℓ) [Fast]Evaluate the validity of this proposed reaction. Justify your answer.The proposed reaction mechanism is as follows:i. BrO3- (aq) + H+ (aq) = HBrO3 (aq) [Fast]ii. HBrO3 (aq) + H+ (aq) = H2BrO3+ (aq) [Medium]iii. H2BrO3+ (aq) + Br- (aq) = Br2O2 (aq) + H2O (ℓ) [Slow]iv. Br2O2 (aq) + 4H+(aq) + 4Br-(aq) = 3Br2 (ℓ) + H2O (ℓ) [Fast]Evaluate the validity of this proposed reaction. Justify your answer.Which of the following mechanism types is/are likely in this reaction: CH3 I + KOH → CH3OH +KI i) SN1 ii) SN2 iii) E1 iv) E2
- An alternative mechanism that may apply when the concentra·tion of O2 is high and that of NO is low is one in which the first step is NO+ O2 → N0···O2, where the dotted line indicates a loosely bound cluster, and its reverse, followed by N0...O2 + NO → NO2 + NO2. Confirm that this mechanism also leads to the observed rate law when the concentration of NO is low.What is the molecularity of the rate limiting step for the mechanism below? Step 1: H2O2 (aq) + I–(aq) → H2O(l) + OI–(aq) (slow)Step 2: H2O2 (aq) + OI–(aq) → H2O(l) + O2 (g) + I–(aq) (fast)The following mechanism is proposed: C6H5COOH ⇌ C6H5CO2H2+ (fast) C6H5CO2H2+ ⇌ C6H5CO+ + H2O (fast) C6H5CO+ + HX + H2O ⟶ C6H6NH2 + N2 + CO2 (slow) The overall enthalpy change is -237.6 kJ. Sketch a reaction-energy diagram consistent with the above mechanism and the enthalpy change. Label the enthalpy change and each transition state. Explain the appearance of your graph.
- The reaction mechanism is given below. elementary step 1 : Br2 ⇋ 2Br fast elementary step 2: H2 + Br à HBr + H slow rate determining step elementary step 3: H + Br2 à HBr + Br fast Overall reaction is H2 + Br2 à 2HBr k1 is the forward reaction rate constant of step 1, k-1 reverse reaction rate constant of step 1, k2 for reaction rate constant of step 2 and k3 for reaction rate constant of step 3. What is the molecularity of step 2? Give the chemical formula of the intermediate(s) in the reaction Deduce the rate law in terms of the reagents for the reaction using the elementary steps.Suppose the formation of nitrosyl chloride proceeds by the following mechanism: step elementary reaction rate constant 1 NO(g)+Cl2(g)->NOCl2(g) k1 2 NOCl2(g)+NO(g)->2NOCl(g) k2 Suppose also k1 ≪ k2. That is, the first step is much slower than the second.The mechanism for the reaction between 2-chloroethanol , CH2CICH2OH, and hydroxide ions in aqueous solution to form ethylene oxide, (CH2CH2)O, is thought to consist of two steps(1) CH2CICH2OH + OH-= CH2CICH2O- + H2O fast(2) CH2CICH2O- = (CH2CH2)O + CI- slowShow that, for this mechanism, the rate of formation of ethylene oxide is Rate= k2 K[CH2CICH2OH][OH-] where K is the equilibrium constant for the first step and k2 is the rate constant for the second step.