CATALYSIS: PARTICIPATIONH2O2(aq) + I−(aq) →H2O(l) + IO−(aq)IO−(aq) + H2O2(aq) →H2O(l) + O2(g) + I−(aq)The mechanism of the decomposition of hydrogen peroxide isWhat is the overall reaction equation?Is there a reaction intermediate? If so, what is it?Is there a catalyst? If so, what is it?What is the rate law for the overall reaction?

Suppose the formation of iodine proceeds by the following mechanism: step elementary reaction 1 H₂(g) + IC1 (g) → HI(g) + HC1 (g) k₁ 2 HI (g) + IC1 (g) → I₂ (g) + HCl (g) k₂ Suppose also k₁ « k₂. That is, the first step is much slower than the second. Write the balanced chemical equation for the overall chemical reaction. Write the experimentally- observable rate law for the overall chemical reaction. Note: your answer should not contain the concentrations of any intermediates. 0 rate = k rate constant

For the decomposition of ammonia on a platinum surface at 856 °C 2 NH3(g) → N2(g) + 3 H2(g) the average rate of disappearance of NH3 over the time period from t = 0 s to t = 6.99x103 s is found to be 1.50x10 6 M/s. What is the average rate of appearance of H2 over the same time period? M/s.

The gas-phase reaction A2(g)+B2(g)→2 AB(g)A2(g)+B2(g)→2 AB(g) is assumed to occur in a single step. Two experiments were done at the same temperature inside rigid containers. The initial partial pressures of A2 and B2 used in experiment 1 were twice the initial pressures used in experiment 2. Which statement provides the best comparison of the initial rate of formation of AB in experiments 1 and 2? a.) The initial rate of formation of AB is the same in both experiments because they were done at the same temperature and the frequency and energy of the collisions between A2 and B2 would have been about the same. b.) The initial rate of formation of AB is slower in experiment 1 than in with experiment 2 because at the same temperature, a higher pressure would reduce the volume available for A2 and B2 molecules to achieve the proper orientation for a successful collision.     c.) The initial rate of formation of AB is faster in experiment 1 than in experiment 2 because at a higher…

Consider the following mechanism for the formation of dinitrogen pentoxide: NO₂(g) + O₂(g) → NO₂(g) + O₂(g) (1) NO₂(g) + NO₂(g) N₂O₂(9) (2) → Write the chemical equation of the overall reaction: Are there any intermediates in this mechanism? If there are intermediates, write down their chemical formulas. Put a comma between each chemical formula, if there's more than one. 0 O yes no

Chlorine dioxide (CO2) is a disinfectant used in municipal water treatment plants. It dissolves in basic solution producing ClO3 and CLO2: 2 ClO2(g) +2 OH(aq) → ClO3(aq) + ClO2 (aq) + H2O(l) The following kinetic data were obtained at 298 K for the reaction. Experiment [ClO2]0 (M) [OH]o (M) Initial Rate (M/s) 1 0.060 0.030 0.0248 2 0.020 0.030 0.00827 3 0.020 0.090 0.0247 Determine the rate law by indicating the values of "x" and "y" in the general rate law below. Rate = k[CO2]*[OH-] x= Select an answer✔ y = Select an answer b. Calculate the rate constant for this reaction at 298 K. Sig. Figs. are important. Rate constant, k = M-1 s-1

Consider the reaction : 3A + 2B →  4C Initially, and at some fixed temperature, the reaction proceeds at such a rate that 2.3 millimoles of A are consumed in 1 minute and 12 secondes. The reaction takes place in 1 L. The following table contains the experimental data for the reaction. (Taken from Ball. Physical Chemistry, 2nd edition, 2006) Rate (M/s) [A], M [B], M 1.081 x 10-5 0.660 1.23 6.577 x 10-5 4.01 1.23 6.568 x 10-5 4.01 2.25   c. Determine the partial order on A. Partial order on B is 0. d. What is the order of the reaction.

Consider the following proposed mechanismHBr + HBr ⇄ H2Br2(fast, equilibrium)HBr + R ⇄ ES(fast, equilibrium)H2Br2 +ES→P+2HBr(slow) (1a) What is the overall reaction? (1b) Identify all catalysts and intermediates in this reaction mechanism. (1c) What is the rate law derived from this mechanism? (1d) If the experimentally determined rate law is rate = k [HBr]3[R]1 what conclusion can be drawn? (1e) If the experimentally determined rate law is rate = k [HBr]2 what conclusion can be drawn? The equilibrium constant (KC) is 10.0 x 10-10 at

The hydrolysis of the sugar sucrose to the sugars glucose and fructose can be described as follows. C12H22011 + H20 → C;H1206 + CgH1206 This reaction follows a first-order rate equation for the disappearance of sucrose: rate = k [C,,H,,0,1] (The products of the reaction, glucose, and fructose, have the same molecular formulas but differ in the arrangement of the atoms in their molecules, i.e. they are 22 isomers.) 1.8x10-11 Arrhenius equation at this temperature). (Enter unrounded values.) (a) In neutral solution, k at 26°C and 8.5x10-11 -1 at 37°C. Determine the activation energy (in kJ/mol), the frequency factor (in s), and the rate constant (in s) for this equation at 48°C (assuming the kinetics remain consistent with the activation energy 108 kJ/mol frequency factor rate constant X s-1 (b) When a solution of sucrose with an initial concentration of 0.130 M reaches equilibrium, the concentration of sucrose is 1.40x10 M. How long (in s) will it take the solution to reach equilibrium…

Suppose the formation of nitrosyl chloride proceeds by the following mechanism: step elementary reaction 1 NO(g) + Cl₂(g) → NOC1₂(9) k₁ 2 NOC1₂(g) + NO(g) → 2 NOCI(g) k₂ Suppose also k₁»k₂. That is, the first step is much faster than the second. Write the balanced chemical equation for the overall chemical reaction: Write the experimentally-observable rate law for the overall chemical reaction. Note: your answer should not contain the concentrations of any intermediates. Express the rate constant k for the overall chemical reaction in terms of k₁, k₂, and (if necessary) the rate constants k_1 and k_2 for the reverse of the two elementary reactions in the mechanism. rate constant rate = k |k=