er 2 800 705 kJ 600 497 kJ PE (kJ) 400 200 175 kJ 100 Reaction Progress Ceated by E Lee for Virtual Virginia (2021) The reaction caardinate diagram above has an activated complex (AC) with a potential energy (PE) of 705 kJ. Imagine that a catalyst is added, bringing the AC PE down to 578 kJ. What will be the new activation energy, in kJ?

er 2 800 705 kJ 600 497 kJ PE (kJ) 400 200 175 kJ 100 Reaction Progress Ceated by E Lee for Virtual Virginia (2021) The reaction caardinate diagram above has an activated complex (AC) with a potential energy (PE) of 705 kJ. Imagine that a catalyst is added, bringing the AC PE down to 578 kJ. What will be the new activation energy, in kJ?

Chemistry: Principles and Reactions

8th Edition

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:William L. Masterton, Cecile N. Hurley

Chapter11: Rate Of Reaction

Section: Chapter Questions

Problem 71QAP: For a certain reaction, Ea is 135 kJ and H=45 kJ. In the presence of a catalyst, the activation...

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For a certain reaction, Ea is 135 kJ and H=45 kJ. In the presence of a catalyst, the activation energy is 39% of that for the uncatalyzed reaction. Draw a diagram similar to Figure 11.14 but instead of showing two activated complexes (two humps) show only one activated complex (i.e., only one hump) for the reaction. What is the activation energy of the uncatalyzed reverse reaction?
Show how equation 20.33 reduces to a simpler form of an integrated first-order rate law when the reverse reaction of an equilibrium is negligible.
The oxidation of iodide ion by the hypochlorite ion in the presence of hydroxide ions I(aq) + ClO(aq) IO(aq) + Cl(aq) was studied at 25 C, and the following initial rates data (Y. Chia and R. E. Connick, Journal of Physical Chemistry, Vol. 63, p. 1518, 1959) were collected: (a) Determine the rate law for this reaction. (b) One mechanism that has been proposed for this reaction is the following: Show that the rate law predicted by this mechanism matches the experimentally determined rate law in part a. (Note that when writing the expression for K the equilibrium constant, [H2O] is not involved. See Chapter 15.)
Experiments have shown that the average frequency of chirping by a snowy tree cricket (Oecanthus fultoni) depends on temperature as shown in the table. Chirping Rate (per min) Temperature (C) 178 25.0 126 20.3 100. 17.3 What is the apparent activation energy of the process that controls the chirping? What is the rate of chirping expected at a temperature of 7.5C?
The equilibrium NH3(aq) + H2O(l) ↔NH4+(aq) + OH−(aq) at 25 °C is subjected to a temperature jump which slightly increases the concentration of NH4+(aq) and OH−(aq). The measured relaxation time is 7.61 ns. The equilibrium constant for the system is 1.78 × 10−5 at 25 °C, and the equilibrium concentration of NH3(aq) is 0.15 mol dm−3. (c) Calculate the rate constant for the reverse reaction. krev = _____________. Just value in 3 sig. fig., in normal or exponential format, e.g. 1.16E6 meaning 1.16 x 106 must use capital E. Choose a unit in the next question. .
he equilibrium NH3(aq) + H2O(l) ↔NH4+(aq) + OH−(aq) at 25 °C is subjected to a temperature jump which slightly increases the concentration of NH4+(aq) and OH−(aq). The measured relaxation time is 7.61 ns. The equilibrium constant for the system is 1.78 × 10−5 at 25 °C, and the equilibrium concentration of NH3(aq) is 0.15 mol dm−3. (a) Calculate the rate constant for the forward step. kf,eff = _____________. Just value in 3 sig. fig., normal or exponential format, e.g. type in 1.16E6 meaning 1.16 x 106, must use capital E here. Choose a unit in the next question, must be in one of those. (b) choose a unit for the forward rate constant. no unit s-1 L/mol/s L2/mol2/s
Determine the activation energy from the following data. Temperature (oC) k (s-1) 288 0.0521 318 0.332
The equilibrium NH3(aq) + H2O(l) ↔NH4+(aq) + OH−(aq) at 25 °C is subjected to a temperature jump which slightly increases the concentration of NH4+(aq) and OH−(aq). The measured relaxation time is 7.61 ns. The equilibrium constant for the system is 1.78 × 10−5 at 25 °C, and the equilibrium concentration of NH3(aq) is 0.15 mol dm−3. (a) Calculate the rate constant for the forward step. kf = _____________. Just value in 3 sig. fig., normal or exponential format, e.g. type in 1.16E6 meaning 1.16 x 106, must use capital E here. Choose a unit in the next question, must be in one of those.