the following data was obtained for the reaction below at 25.0°C 2 NO(g) + O2(g) --> 2 NO2(g) (see image) Determine the rate law for this reaction and calculate the rate constant k including the correct units
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the following data was obtained for the reaction below at 25.0°C
2 NO(g) + O2(g) --> 2 NO2(g)
(see image)
Determine the rate law for this reaction and calculate the rate constant k including the correct units
![Trial
1
2
3
Initial Reactant Concentration (M)
[0₂]
0.0055
0.0110
0.0055
[NO]
0.030
0.030
0.060
Initial Rate (M/s)
8.55 × 10-3
1.71 x 10-2
3.42 x 10-2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1385b10f-f9b1-451c-97aa-8152fc0259bb%2Faac14494-fe0f-4486-aa01-631a66b53bdc%2Fj6pibvp_processed.jpeg&w=3840&q=75)
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- NH4+ {aq) + NO2(aq) -> N2(g) +2H2O{l} Data Initial [NH4+] Initial [NO2-] rate 1 0.0100 0.200 5.4 x10-7 2 0.0200 0.200 10.8x10-7 3 0.0400 0.200 21.5x10-7 4 0.200 0.0202 10.8x10-7 5 0.200 0.0404 21.6x10-7 6 0.200 0.0808 43.3x10-7 Find x,y,kExpt. 1 2 3 NH; (aq) + NO; (aq) >Nz (aq) + 2 H2O (l) [NO₂] Rate (at 298 K) 0.020 M/s 0.020 M 0.020 M 0.030 M/s 0.010 M 0.0050 M/s [NH+] 0.010 M 0.015 M 0.010 M Determine the complete rate law (rate law and rate constant) for the reaction. Show all work. What is the initial rate of this reaction if the concentrations in Expt. 2 above are tripled. Show work. What is the rate of H₂O formation in Expt. 1? A 4th trial is conducted at 273 K, using the concentrations from Expt. 1. If the rate was found to be 0.016 M/s, calculate the activation energy of this reaction.Calculation Tables Table 4- Logarithms [l]o (M) -log[I]o -log(rate) [H2O2]o (M) 0.0100 -log[H2O2]o 2.00 Experiment # Rate (M/s) 1 2.63 x10-6 5.58 0.0200 1.70 5.18 x 106 0.0100 0.0400 0.0600 8.01 x 10-6 1.62 x 105 0.0100 4 0.0200 0.0600 3.21 x 105 0.0400 0.0600 Table 5- Rate Constants Experiment [H2OzJo(M) [H2O2]om Initial rate k # (M/s) 2.63 x10-6 1 0.0100 0.0200 2 0.0100 0.0400 5.18 x 10-6 0.0100 0.0600 8.01 x 106 4 0.0200 0.0600 1.62 x 105 5 0.0400 0.0600 3.21 x 105 Value of m (2 significant figures): Value of n (2 significant figures): verage value of k (3 significant figures):
- just answer it on a sheet of paperA student completed the lab and obtained the following concentration vs rate data: Trial [CV*L (M) [OH-1L(M) Rate 1st 30 sec(M/s) 1.00E-05 5.00E-03 6.290E-09 4.00E-06 5.00E-03 2.610E-09 3 1.00E-05 1.00E-03 1.320E-09 Based on their data, complete the following regarding the rate order dependence and rate law. Give your answer as a decimal to the nearest hundredth. [CV*] (OH'] Rate=k[CV*] [OH']Fast pls solve this question correctly in 5 min pls I will give u like for sure Mass of Copper: 0.99g Mass of Aluminum: 0.35g Mass of CuCl2: 2.55g 2Al(s) + 3CuCl2 * 2H2O (Hydrate molecule) --→ AlCl3(aq) + 3Cu + 6H2O Please calculate the limiting reagent and show the steps.
- The following data was collected for the reaction between ethoxide and methyl iodide T(°C) Rate (1/s) T (K) 1/T (1/K) In (rate) 168 273.15 0.00366 5.12 6 354 279.15 0.00358 5.87 12 735 285.15 0.00351 6.60 18 1463 291.15 0.00343 7.29 24 3010 297.15 0.00337 8.01 30 6250 303.15 0.00330 8.74 A plot of In (rate) vs. 1/T using the above table was made and is shown below, and a linear regression line fitted to the data. Using the provided graph, estimate the activation energy (kJ/mol) of the reaction between ethoxide and methyl iodide. Enter your number using 2 decimal places without unit. Gas constant is R = 8.3144598 J. mol 1. K-1. In(k) vs 1/T 9.30 8.80 8.30 7.80 y = -9930.5x + 41.445 R² = 0.9992 2 7.30 6.80 6.30 5.80 5.30 4.80 0.00325 0.00330 0.00335 0.00340 0.00345 0.00350 0.00355 0.00360 0.00365 0.00370 1/T (1/K) Ln(k)Answer both questions at bottomIV. DOER: Now do it! Provided the reaction A + 2B + C → 2D + E, and the set of data that follows: Experiment rate no. 1 1.4 1.4 1.00 R1 0.7 1.4 1.00 R1 R2 = 3 0.7 0.7 1.00 R2 R3 = 1.4 1.4 0.50 R4 = 16R3 0.7 0.7 0.50 wherein the rate law is hypothetically identified as rate = k [A]* [B]Y [CF a. what are the reaction orders with respect to A, B, and C? b. what is the value of Rs in terms of the variable R1? C. what is rate law for the above reaction? C. 2.
- the mean time, the rate, [S₂O3] in the reaction may label clear outliers the mean) as such and e calculation. on mixtures NH4)2SO4 (NH4)2S2O8| (mL) (mL) 1.50 0.50 1.50 0.50 1.00 1.00 1.00 1.00 following reagents are um iodide ium persulfate m thiosulfate um nitrate ium sulfate odide in the solution in the Answer: 0.050 M • Determination of the rate law follows the protocol in ref. (5). with one difference. A reactant order (rate law exponent) will be calculated twice using two different pairs of determinations (e.g. determinations 1 & 2 and determinations 3 & 4). Then the mean is taken of both values and reported as order. This is done for each reactant. This additional step allows us to include more data and reduce experimental error. Practice Problem 2: The rate of a reaction CO(g) +3H₂(g)→ CH4(g) + H₂O(g) was determined at 25 °C. From the following data, determine a) the mean reaction order with respect to CO b) the mean reaction order with respect to H₂ c) the overall rate order d)…Can you tell me where did i go wrong for the experiment 1 and 2?3. The reaction H₂SO3(aq) + 6(aq) + Haq) → Se(s) +213(aq) + 3H₂0 (1) was studied at 0°C. The following data were obtained: Experiment # 1 2 3 4 5 6 7 [H₂SO3(aq)lo (mol/L) constant? 1.0 x 104 2.0 x 10-4 3.0 x 104 1.0 x 10-4 1.0 x 10-4 1.0 x 104 1.0 x 104 [¹] (mol/L) 2.0 x 10² 2.0 x 10-² 2.0 x 10-² 4.0 x 10-² 1.0 x 10-² 2.0 x 10-² 1.0 x 10-² Find the order of the reaction with respect to [H₂SO3]. respect to [I-¹]. reaction with respect to [H+¹]. is the overall order of the reaction? Find the value of the rate constant (k). Only write the number. [H¹+] (mol/L) 2.0 x 10-² 2.0 x 10-² 2.0 x 10-² 2.0 x 10-² 2.0 x 10-² 4.0 x 10-² 4.0 x 10-² Rate (mol/L.s) Find the order of the reaction with Find the order of the What What are the units of the rate 1.66 x 10-7 3.33 x 10-7 4.99 x 10¹7 6.66 x 10-7 0.42 x 10-7 13.2 x 10-7 3.36 x 10-7