Plotting Reaction Progression: In each case make reasonable, relative, plots of every rea every product in relation to each other as a function of time. Balance first and think about the effect of a limiting reagent 14. NO, + Cl → NO;Cl [NO:], = [Cl:], = 0.6 M
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- For the following reaction 6 experiments have been run and the data collected is in the following table: 2 MnO4-(aq) + 5 H2C2O4 (aq) + 6 H+ (aq) ---> 2 Mn2+ (aq) + 10 CO2 (g) + 8 H2O (l) Experiment [MnO4-], M [H2C2O4], M [H+], M Rate, M/s 1 0.2410 0.3470 0.2690 0.1147 2 0.3260 0.6210 0.2270 0.2776 3 0.5630 0.5740 0.7420 0.4431 4 0.2410 0.3470 0.3840 0.1147 5 0.4140 0.5740 0.5610 0.3258 6 0.3260 0.4930 0.4910 0.2203 a) Find the order of the reaction with respect to H+. _______________ b) Find the order of the reaction with respect to MnO4-. ______________ c) Find the order of the reaction with respect to H2C2O4. _____________ d) What is the overall order of the reaction? ________________________b) what is the velocity of a reaction when [S] = 20 mM, where Vmax = 100 umol/mL*sec, and 4 Km = 2mM? can you show me how he got 98 umol / mL*sec, very close to Vmx, as [S]>>Km, at zero order of the kinetics. step by step pleaseReaction of interest : S2O82-(aq) + 3I- (aq)→ 2SO42-(aq) + I3-(aq). rate= k[S2O82-]1[I-]1.
- For the reaction 2N2O5(g) → 4NO2(g) + O2(g), the following data were collected. t (minutes) [N2O5] (mol/L) 0 1.24 × 10–2 10. 0.92 × 10–2 20. 0.68 × 10–2 30. 0.50 × 10–2 40. 0.37 × 10–2 50. 0.28 × 10–2 70. 0.15 × 10–2 The concentration of O2 at t = 10. min is Question 16 options: A) 0.32 × 10–2 mol/L B) 2.0 × 10–4 mol/L C) 0.64 × 10–2 mol/L D) 0.16 × 10–2 mol/L E) none of theseConsider the reaction A(g) + 2B(g) → 2C(g) + 2D(g) and associated that was collected at a constant temperature. Trial Initial [A] Initial [B] Initial Rate (mol/L) (mol/L) (mol/(L·min)) 1 0.125 0.200 7.25 2 0.375 0.200 21.75 3 0.250 0.400 14.50 4 0.375 0.400 21.75The following reaction is carried out with [PCl5]i = 0.400 M at 25°C (i means initial) The reaction: PCl5(g)⇌PCl3(g)+Cl2(g) K= 0.00105 Using the small x approximation and "test x", could you please find the percent to two decimal places when testing x? Thank you so much.
- how do i find the time it took for the mixture to react from A = 0.600 to 0.800 when catalyst was added to the reactant?Suppose that for the reaction K + L------->M, you monitorthe production of M over time, and then plot the followinggraph from your data: (a) Is the reaction occurring at a constant rate from t = 0to t = 15 min? (b) Is the reaction completed at t = 15min? (c) Suppose the reaction as plotted here were startedwith 0.20 mol K and 0.40 mol L. After 30 min, an additional0.20 mol K are added to the reaction mixture.Which of the following correctly describes how the plotwould look from t = 30 min to t = 60 min? (i) [M] wouldremain at the same constant value it has at t = 30 min,(ii) [M] would increase with the same slope as t = 0 to15 min, until t = 45 min at which point the plot becomeshorizontal again, or (iii) [M] decreases and reaches0 at t = 45 min.For the reaction A---> B the dissapearance of A is first-roder where k-0.030/min. If we begin with [A]=0.36 M, what will [A] be after 46 min?
- For the reaction 2N2O5(g) → 4NO2(g) + O2(g), the following data were collected. t (minutes) [N2O5] (mol/L) 0 1.24 × 10–2 10. 0.92 × 10–2 20. 0.68 × 10–2 30. 0.50 × 10–2 40. 0.37 × 10–2 50. 0.28 × 10–2 70. 0.15 × 10–2 The concentration N2O5 at 100 min will be approximately Question 17 options: A) 0.10 × 10–2 mol/L B) 0.01 × 10–2 mol/L C) 0.06 × 10–2 mol/L D) 0.03 × 10–2 mol/L E) none of theseConsider the experimental data collected for the following chemical reaction: 2H2(g) + 2NO(g) → N2(g) + 2H2O(g) Trial Initial [H2] (mol/L) Initial [NO] (mol/L) Rate of Reaction (mol/L∙s) 1 0.031 0.0025 2.0 x 10-3 2 0.062 0.0025 4.0 x 10-3 3 0.031 0.0050 8.0 x 10-3 4 0.062 0.0050 1.6 x 10-2 Which action will have more effect on the rate of this reaction; changing the concentration of H2(g) or that of NO(g)? Explain your reasoning by considering the orders of reactants and writing the corresponding rate law equation.Consider the following organic reaction, in which onehalogen replaces another in an alkyl halide:CH₃+CH₂+BrKI→CH₃+CH₂I+KBr .In acetone, this particular reaction goes to completion becauseKI is soluble in acetone but KBr is not. In the mechanism, Iapproaches the carbon oppositeto the Br . After Br⁻ has been replaced by I⁻ and precipitates as KBr, other I⁻ ions react with the ethyliodide by the same mechanism.(a) If we designate the carbon bonded to the halogen as C-1,what is the shape around C-1 and the hybridization of C-1 inethyl iodide?(b) In the transition state, one of the two lobes of the unhy-bridized 2porbital of C-1 overlaps a porbital of I, while theother lobe overlaps a porbital of Br. What is the shape aroundC-1 and the hybridization of C-1 in the transition state?(c) The deuterated reactant, CH₃CHDBr (where D is deuterium,2H), has two optical isomers because C-1 is chiral. If the reac-tion is run with one of the isomers, the ethyl iodide is notopti-cally active.…