What is the rate law expression for the reaction in sodium hydroxide and ethyl acetate? Trial NaOH CH3COOC2H5 Initial Condcutivity of NaOH Solution (uS/cm) Initial Rate (uS/cm)/s) 1 21 ml 2.1ml 1285 24.72 2 21 ml 2.1ml 2399 50.32 3 21 ml 2.1ml 1262 23.72
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- What is the rate law expression for the reaction in sodium hydroxide and ethyl acetate?
|
Trial |
NaOH |
CH3COOC2H5 |
Initial Condcutivity of NaOH Solution (uS/cm) |
Initial Rate |
|
1 |
21 ml |
2.1ml |
1285 |
24.72 |
|
2 |
21 ml |
2.1ml |
2399 |
50.32 |
|
3 |
21 ml |
2.1ml |
1262 |
23.72 |
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- Table 2: Molarity of H2O2 and KI and Reaction Rate Trial H2O2 Concentration, M KI Concentration, M Reaction Rate(Reciprocal Slope) 1 0.29 M 0.40 M 14.08 2 0.29 M 0.20M 25 3 0.023 M 0.40 M 20 Please help me with this part, and please show me all steps Order with respect to H2O2: Order with respect to KI:Table 2: Molarity of H2O2 and KI and Reaction Rate Trial H2O2 Concentration, M KI Concentration, M Reaction Rate(Reciprocal Slope) 1 0.29 M 0.40 M 14.08 2 0.29 M 0.20M 25 3 0.023 M 0.40 M 20 help me with this part please, and show me all steps Order with respect to H2O2: Order with respect to KI:Table 2: Molarity of H2O2 and KI and Reaction Rate Trial H2O2 Concentration, M KI Concentration, M Reaction Rate(Reciprocal Slope) 1 0.29 M 0.40 M 14.08 2 0.29 M 0.20M 25 3 0.023 M 0.40 M 20 Order with respect to H2O2: -0.138 Order with respect to KI: -0.828 Rate constant value Trial 1: 5.84 Trial 2: 5.55 Trial 3: 5.56 Average: 5.6 -Please help me with this part, please 1. Should the rate constant (k) be the same for all three trials in this experiment? Explain your answer. 2.Write the complete rate law for this reaction.
- The kinetic data shown below were observed for the reaction: BF3 (g) + NH3 (g) ---------> F3B·NH3 (g) Trial # [BF3] (mol/L) [NH3] (mol/L) Rate (M/s) 1 0.250 0.250 0.2130 2 0.250 0.125 0.1065 3 0.200 0.100 0.0682 4 0.350 0.100 0.1193 5 0.175 0.100 0.0596 What is the rate constant for this experiment?Derive a rate law for the following proposed mechanisms and determine if it is consistent with the observed rate law. Reaction: H2+Br2 --> 2 HBr Observed Rate Law for Reaction: rate = k [H2][Br2]1/2 Proposed Mechanism: H2 ⇌ 2 H fast k1 for foward reaction, k1 for reverse reaction H + Br2 --> H Br + Br slow k2 for foward reaction Br + H --> H Br fast k3 for foward reactionTable 2: Molarity of H2O2 and KI and Reaction Rate Trial H2O2 Concentration, M KI Concentration, M Reaction Rate(Reciprocal Slope) 1 0.29 M 0.40 M 14.08 2 0.29 M 0.20M 25 3 0.023 M 0.40 M 20 Rate constant value Trial 1 ______________ Trial 2 ______________ Trial 3 ______________ Average _____________
- Derive a rate law for the following proposed mechanisms and determine if it is consistent with the observed rate law. Reaction: H2+Br2 --> 2 HBr Observed Rate Law for Reaction: rate = k [H2][Br2]1/2 Proposed Mechanism: Br2 ⇌ 2 Br fast k1 for foward reaction, k1 for reverse reaction Br + H2 --> H Br + H slow k2 for foward reaction H + Br --> H Br fast k3 for foward reactionThe following results were found after completion of Part C in the Experimental procedure: 0.057 M I- and 0.065 M H2O2 were used Run Catalyst Calculated Rate of Reaction (M/s) 1 none 0.053 2 10.00mL of 0.54 M FeCl2 0.446 Assuming the Rate Law = k[I-][H2O2] 2 Calculate the value of k for run 2. Give your answer to the nearest whole number.Calculate the full rate law using the information below: 2 NACHOS (g) → 12 CHIPS (g) + 3 CHEESE (g) [NACHOS]0 (M) Initial Rate (M–1s–1) 0.101 0.00484 0.0910 0.00437 Group of answer choices Rate = 5.61 × 10–2 M–1 s–1 [NACHOS]2 Rate = 6.03 × 10–1 M–2 s–1 [NACHOS]3 Rate = 4.79 × 10–2 s–1 [NACHOS] Rate = 1.66 × 10–3 M1/2 s–1 [NACHOS]2/3 Rate = 1.78 × 10–2 M–1/2 s–1 [NACHOS]3/2
- Determine the missing initial rate: 2N2O5 → 4NO2 + O2 [N2O5] Initial rate (Ms-1) 0.093 4.84x10-4 0.084 4.37x10-4 0.224 ??The recombination reaction2 HO2(g) → H2O2(g) + O2(g)has a second-order rate constant of 1.7E9 M-1 s-1. Calculate the half-life of the reaction if the initial concentration of HO2(g) is 2.22E-11 M.Note: the coefficient 2 in the reaction means that the integrated rate law for this reaction is:1/ct = 1/c0 + 2kt Give the answer to 3 significant figuresUsing the data in the table, determine the rate constant of the reaction and select the appropriate units. A+2B⟶C+D Trial [?] (?) [?] (?) Rate (M/s) 1 0.340 0.210 0.0204 2 0.340 0.420 0.0204 3 0.680 0.210 0.0816 ?=