Q:4-2The iodide ion reacts with hypochlorite ion (the ac-tive ingredient in chlorine bleaches) in the following way:OCI¯ + I¯ –→ 0I¯ + Cl¯. This rapid reaction gives thefollowing rate data:[OCI-] (M)[1-] (M)Initial Rate (M/s)1.5 × 10-31.5 × 10–31.36 × 10¬43.0 × 10–31.5 × 10–32.72 × 10¬41.5 × 10¬33.0 × 10–32.72 X 10¬4(a) Write the rate law for this reaction. (b) Calculate therate constant with proper units. (c) Calculate the rate when[OCr] = 2.0 × 10¯³ M and [I¯] = 5.0 × 10-ª M.[OCF]

(a) Let order of reaction is x with respect to OCl- and y with respect to I- . Rate law is: r = k…

The iodide ion reacts with hypochlorite ion (the ac- tive ingredient in chlorine bleaches) in the following way: OCI- + I- –→ OI¯ + CI¯. This rapid reaction gives the following rate data: [OCI-] (M) [I-] (M) Initial Rate (M/s) 1.5 X 10-3 1.5 × 10-3 1.36 × 104 3.0 x 10-3 1.5 × 10-3 2.72 × 10¬4 1.5 X 10-3 3.0 × 10-3 2.72 X 10¬4 (a) Write the rate law for this reaction. (b) Calculate the rate constant with proper units. (c) Calculate the rate when [OCF] = 2.0 x 10³ M and [I¯] = 5.0 × 10¬ª M.

The iodide ion reacts with hypochlorite ion (the ac- tive ingredient in chlorine bleaches) in the following way: OCI- + I- –→ OI¯ + CI¯. This rapid reaction gives the following rate data: [OCI-] (M) [I-] (M) Initial Rate (M/s) 1.5 X 10-3 1.5 × 10-3 1.36 × 104 3.0 x 10-3 1.5 × 10-3 2.72 × 10¬4 1.5 X 10-3 3.0 × 10-3 2.72 X 10¬4 (a) Write the rate law for this reaction. (b) Calculate the rate constant with proper units. (c) Calculate the rate when [OCF] = 2.0 x 10³ M and [I¯] = 5.0 × 10¬ª M.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

NO + 03 (9) NO2(g) + O2 (g) This reaction has been studied in the laboratory, and the following rate data were determined at 25 °C. Trial [NO] (mol/L) [03] (mol/L) Delta [NO,] (molL-s) At 1.00 x 106 3.00 x 105 6.60 x 10 5 100 x 10 6 6.00 x 106 1.32 x 104 1.00 x 106 9.00 x 10 6 1.98 x 104 2.00 x 106 9.00 x 10 6 3.96 x 10 4 5. 3.00 x 10 6 9.00 x 106 5.94 x 104 Determine the rate law and the rate constant for the reaction at 25 °C. 3.
Using the data in the table, determine the rate constant of the Trial [A] (M) [B] (M) Rate (M/s) reaction and select the appropriate units. 1 0.390 0.260 0.0211 0.390 0.520 0.0211 А + 2B — С + D 3 0.780 0.260 0.0844 Units k =
At elevated temperatures, dinitrogen pentoxide decomposes to nitrogen dioxide and oxygen: 2N205(8)→ 4NO2 (g) + O2(g) When the rate of formation of O2 is 3.6 x 10-3 M/s, the rate of decomposition of N2O5 is. M/s. A) 0.00090 OB) 1.4 x 10-2 1.8 × 10-3 OD) 3.6 × 10-3 E) 4.4 x 10-4
[References] The reaction I (ag) + OCI (aq) → IO¯(aq) + Cl¯(aq) was studied, and the following data were obtained: I], (mol/L) [OCI"], (mol/L) Initial Rate (mol/L · s) 0.12 0.16 7.80 x 10-2 0.060 0.16 3.90 x 10-2 0.030 0.080 9.80 x 10-3 0.24 0.080 7.80 x 10- What is the rate law? (Use k for the rate constant.) Rate Calculate the value of the rate constant. k = L/mol · s Calculate the initial rate for an experiment where both I and OcI are initially present at 0.16 mol/L. Rate = mol/L · s
3. What is the rate of reaction with respect to HI(g) in the following reaction: H₂(g) + I2(g) → 2HI(g) (a) 2 [HI] dt 1 d[HI] (b) (c) 2 dt 1 d[HI] 2 dt (d) -2. dt d[HI]
the chemistry of smog formation includes NO3 as an intermediate in several reactions. If Δ[NO3] / Δt is –21.58×104 mM/min in the following reaction, what is the rate of appearance of NO2? NO3(g)+NO(g) ->2NO2(g) Δ[NO2]Δt= mM/min
If the rate law for a reaction is rate=k[A]2[B], a). If the concentration of A is doubled and the concentration of B is tripled, how will this affect the rate of the reaction? b).By what factor will the rate of the reaction change? (enter 0 if the rate will not change)
I need help please. There's 3 problems as shown in the second picture. In each one the reactant and the product are given. The reactant is on the left and the product is on the right. I need help figuring out how to get from the reactant to the product. Please show the steps on how to get to the product. I attached a picture of how I solved some, please follow the same pattern.
2 NO(g) + Br2 (g) → 2 NOB1(g) Initial Rate Experiment NO] (M) [Br2] (M) (mol/L's) 1 1.0 x 10-2 2.0 × 10-2 2.4 x 10-2 2 4.0 x 10-2 2.0 × 10-2 0.384 3 1.0 x 10-2 6.0 x 10-2 7.2 x 10-2 What is the order of the reaction with respect to [NO] and [Br2], and what is the overall order of the reaction? Order with respect to [NO] = Order with respect to [Br2] = Overall order
Examine the 2-step mechanism below and identify the substance that appears to be a catalyst and anintermediate, respectively.Step 1. A + BC → AC + BStep 2. AC + D → A + CD(a) A = catalyst, AC = intermediate(b) AC = catalyst, A = intermediate(c) A = catalyst, B = intermediate(d) D = catalyst, AC = intermediate
a- 3A +2BP (one step) write the rate equation b- If [A] is the rate 2x and [B] is 2x what happens to
Enter your answer in the provided box. Consider the reaction: 2NO(g) + 0,(g) –→ 2NO,(g) Suppose that a particular moment during the reaction nitric oxide (NO) is reacting at a rate of 0.062 M/s. At what rate is molecular oxygen reacting? M Molecular oxygen is (select) V at a rate of