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- The color change accompanying the reaction of phenolphthalein with strong base is illustrated below. The change in concentration of the dye can be followed by spectrophotometry (Section 4.9), and some data collected by that approach are given below. The initial concentrations were [phenolphthalein] = 0.0050 mol/L and [OH] = 0.61 mol/L. (Data are taken from review materials for kinetics at chemed.chem.purdue.edu.) (For more details on this reaction see L Nicholson, Journal of Chemical Education, Vol. 66, p. 725, 1989.) (a) Plot the data above as [phenolphthalein] versus time, and determine the average rate from t = 0 to t = 15 seconds and from t = 100 seconds to t = 125 seconds. Does the rate change? If so, why? (b) Use a graphical method to determine the order of the reaction with respect to phenolphthalein. Write the rate law, and determine the rate constant. (c) What is the half-life for the reaction?Instantaneous rates for the reaction of hydroxide ion with Cv+ can be determined from the slope of the curve in Figure 11.3 at various concentrations. They are (1) At 4.0 105 mol/L, rate = 12.3 107 mol L1 s1 (2) At 3.0 105 mol/L, rate = 9.25 107 mol L1 s1 (3) At 2.0 105 mol/L, rate = 6.16 107 mol L1 s1 (4) At 1.5 105 mol/L, rate = 4.60 107 mol L1 s1 (5) At 1.0 105 mol/L, rate = 3.09 107 mol L1 s1 (a) What is the relationship between the rates in (1) and (3)? Between (2) and (4)? Between (3) and (5)? (b) What is the relationship between the concentrations in each of these cases? (c) Is the rate of the reaction proportional to the concentration of Cv+? Explain your answer.First order: [A]o In———- = kt [A]t t1/2= .693/k
- Q4. lodide ions are oxidised to iodine by hydrogen peroxide in acidic conditions. H₂O₂(aq) + 2H+(aq) + 21-(aq) → 1₂(aq) + 2H₂O(1) The rate equation for this reaction can be written as rate = k [H₂O₂] [1] [H*]c In an experiment to determine the order with respect to H*(aq), a reaction mixture is made containing H*(aq) with a concentration of 0.500 mol dm-³ A large excess of both H₂O2 and 1 is used in this reaction mixture so that the rate equation can be simplified to rate = k₁ [H*]c (a) Explain why the use of a large excess of H₂O2 and I means that the rate of reaction at a fixed temperature depends only on the concentration of H+(aq). (b) Samples of the reaction mixture are removed at timed intervals and titrated with alkali to determine the concentration of H*(aq). State and explain what must be done to each sample before it is titrated with alkali.The gas phase reaction between hydrogen and iodine is proposed to occur as follows:.....step 1.....fast:......I2 2 I.....step 2.....slow:....H2 + 2 I 2 HI(1) What is the equation for the overall reaction? Use the smallest integer coefficients possible. If a box is not needed, leave it blank. + + (2) Enter the formula of any species that acts as a reaction intermediate? If none leave box blank: (3) Complete the rate law for the overall reaction that is consistent with this mechanism.Use the form k[A]m[B]n... , where '1' is understood (so don't write it if it's a '1') for m, n etc. Rate =Hydrogen peroxide and the iodide ion react in acidic solution as follows: H2O2 (ag) + 31 (aq) + 2H† (ag) → I3 (aq) + 2H2O(1) The kinetics of this reaction were studied by following the decay of the concentration of H2O2 and constructing plots of In[H2 O2] versus time. All the plots were linear and all solutions had [H2O2], = 8.0 x 10-4 mol/L. The slopes of these straight lines depended on the initial concentrations of I¯ and H+. The results follow: I], (mol/L) [H*], (mol/L) Slope (min) 0.1000 0.0400 -0.120 0.3000 0.0400 -0.360 0.4000 0.0400 -0.480 0.0750 0.0200 -0.0760 0.0750 0.0800 -0.118 0.0750 0.1600 -0.174 The rate law for this reaction has the form -d[H2O2] (kı + k2 [H*])[I¯]™ [H2O2]" Rate = dt a Specify the orders of this reaction with respect to [H2O2] and [I¯]. п %3 т —
- 86) 203 (9) 7 302 (9) What is the Rate Law? 1.0L vessel => in which 10 mole * 1.0 mole 03 * 1.0 mole 02 What fraction of 03 will have reacted when the rate falls to one-fourth of its initial valuc?2. a) Calculate the hard-sphere (HS) collision theory rate constant ks for the reaction NO(g) + C₁₂(g) Þ NOCl(g) + HS Cl(g) at 300 K in units of dm³ mols. The collision diameters of NO and Cl₂ are 370 and 540 pm, respectively. 16.0 g/mol, and mcı Use masses of m₁ = 14.0 g/mol, mo = 35.0 g/mol. N = b) Now include the line-of-centers (LOC) energy criterion to find koc in units of dm³ mol¹ s¹ using E₁ = 84.9 kJ mol LOC LOC c) Calculate the ratio of kчs and kLoc to the experimental rate constant at 300 K given A = 3.981 10° dm³ mol¹ s and E₁ = 84.9 kJ mol¹. a -1Give the equation for quantum yield j. rate of formation of the singlet state rate of ith decay mechanism rate of ith decay mechanism rate of removal of the singlet state rate of ith decay mechanism rate of formation of the singlet state rate of removal of the singlet state rate of ith decay mechanism Step 4 of 6 Give the quantum yield ;. (Use the following as necessary: kj, k1, k2, k3, and t.) k1 + k, + k = ki 3 k1 + k2 + k3 Recalling the relationship between the overall rate constant, k, and the mean lifetime, T, give the quantum yield D; in terms of T and Tj. (Use the following as necessary: T and T;) i Step 5 of 6 (c) If T1 = 1×10-7 s, T2 = 4x10-8 s, and T3 = 1x10-8 s, calculate the lifetime of the singlet state and the quantum yield for the path that has T2. (Enter an unrounded value with at least 3 digits.) 1 using the equation given in (b). Calculate = 7.407e-9 s-1 Solve for T, the lifetime of the singlet state. T = .185175 X S Submit
- The numerical value of the term e-kt at a time equalto 3 half lives of a drug will always be equal toa. 0.5b. 0.75c. 0.05d. 0.125 How long will it take to lose 75% of the administered doseof a drug from the body if the drug follows first orderprocess?a. 3,32 half livesb. 2,32 half livesc. 2.0 half livesd. None of the aboveThe slope of an Arrhenius plot was found to be -4570 K, and the intercept was found to be 3.3 with the time plotted in seconds. What is the frequency factor? Enter your answer with two sig figs.In a study of the gas phase decomposition of dimethyl ether at 500 °C CH3OCH3(gCH,(g) + H2(g) + CO(g) the concentration of CH3OCH3 was followed as a function of time. It was found that a graph of In[CH3OCH3] versus time in seconds gave a straight line with a slope of -6.08×104 s and a y-intercept of -2.39 . Based on this plot, the reaction is | order in CH3OCH3 and the rate constant for the reaction is