Determine the rate law and the value of k for the following reaction using the data provided. 2 NO(g) + Br2(g) → 2 NOBr(g) [NO]i (M) [Br2]i (M) Initial Rate (Ms-1) 0.030 0.0055 1.81 x 10-2 0.030 0.0110 3.62 x 10-2 0.060 0.0055 7.25 x 10-2 |
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Determine the rate law and the value of k for the following reaction using the data...
Part A Determine the rate law and the value of k for the following reaction using the data provided: 2NO(g) + O2(g) + 2NO2(9) [NO]; (M) [Ogl (M) Initial Rate (M4 st) 0.030 0.0055 8.55 x 103 0.030 0.0110 1.71 x 10-2 0.0055 3.42 x 10-2 0.060 Rate = 3.1 x 105 M-3 s[NO]2[0212 Rate = 57 M' s'[NO][O21 Rate = 9.4 x 103 M25 *[NO][0212 Rate = 3.8 M-1/2 s [NO](O2) 1/2 Rate = 1.7 x 103 M2 s?[NO]2[02]
PW 5) Consider the following reaction and the data of initial rates measured using different concentrations of reactants. [10 pt) 2 NO(g) + O2(g) → 2 NO2(g) [NO]i (M) 0.030 0.030 0.060 [O2li (M) 0.0055 0.0110 0.0055 Initial Rate (Mºs-1) 8.55 x 10-3 1.71 x 10-2 3.42 x 10-2 a) Determine the reaction order of the chemical reaction. [5 pt] b) Calculate the rate constant (k) including the unit and write the rate law of the reaction. TS 1
37. The data below were determined for the reaction shown below. The rate law for this reaction 12 Marks must be: s2or" + 31-(aq)-25042-+13- Initial Rate 1.4 × 10-5 M/s 2.8 × 10-5 MS 1.4 x 10-5M/s 0.060 0.060 0.030 0.038 0.076 0.076 2 a. Rate kIS 0s I] b. Rate kIS20s IUT *d. Rate = k[S2082-11
Determine the rate law and the value of k for the following reaction using the data provided. 2 N2O5(g) → 4 NO2(g) + O2(g) [ N2O5]i (M) Initial Rate (M-1s-1) 0.033 1.84 × 10-4 0.066 7.37 × 10-4 0.099 1.66× 10-3 A. Rate = 5.6 × 10-1 M-1/2s-1[N2O5]3/2 B. Rate = 1.7 × 10-1 M-1s-1[N2O5]2 C. Rate = 1.7 × 10-1M1/2s-1[N2O5]1/2 D. Rate = 6.0 × 10-1 M-2s-1[N2O5]3 E. Rate = 5.2 × 10⁻1 s-1[N2O5]
The reaction 2NOB → 2NO+ Br2 exhibits the rate law Rate = k[NOBr] = A[NOBr) A: where k = 1.0 x 10-5 M-1.5 at 25° C. This reaction is run where the initial concentration of NOBT ([NOBr]) is 1.00 x 10- M. Calculate the [NO] (M) after 3.6 hours have passed. Report your answer to two significant figures. Answer:
Question Determine the rate law and the value of k for the following reaction using the data provided. 1/1 point 2 N2O5 - 4 NO2+O IN20sl (M) initial Rate (M-1,-1) 0.093 4.84 x 10-4 0.084 4.37 x 10-4 0.224 1.16 x 10-3 Rate - 6,0 x 10-1M2-1N205) Rate -5.2 x 10-3541N205) Rate - 5.6 x 10 2 M15-41N20512 Rate - 1.6 x 10-3 M1/2,-41N20511/2 Rate - 1.7* 10-2M-1/2--IN20513/2
Show work please. So I know how to apporach problems like these in the future. (2 points) Determine the complete rate law for the following reaction using the data provided. 7. 2 NO(g) +O2(g)2 NO2(g) [NO]i (M) [O2]i (M) Initial Rate (M-Is-I) 0.030 0.030 0.060 0.00558.55 x 10-3 0.01101.71 x 10-2 0.0055 3.42×10-2 What is the initial rate if the concentration of both species is 0.050 M? (1 points) Consider the following reaction: A reaction mixture initially contains 2.24 atm...
Homework 8-3_1 Reaction Rate Orders, Temp Effects Reaction Rates and Rate Orders The following initial rate data was collected for the reaction: HI(g) + CHşl(g) à CH.(g) +1-(9) Experiment (HI) (CHI) Initial Rate 0.015 M0.900 M 4.01 x 10M's 0.030 M0.900 M 8.04 x 10'Ms 0.030 M0.450 M 3.99 x 10 m/s 1. What is the rate law for this reaction? a. Rate =k [HI] [CHI] b. Rate = k [HI] [C,H,I] c. Rate = k [HI] (C,H,I] d. Rate...
Consider this initial-rate data at a certain temperature for the reaction described by [SO, Cl, 1, (M) Initial rate (M/s) SO,Cl2(g) → SO2(g) + Cl2(g) 1.80 x 10-6 Determine the order with respect to SOC12. 0.100 0.200 0.300 3.60 x 10-6 5.40 x 10-6 order: Determine the value and units of the rate constant. units: Initial rate data at a certain temperature is given in the table for the following reaction. 2 NOBr(g) —> 2 NO(g) + Br2(g) [NOBr), (M)...
Part A Determine the rate law and the value of k for the following reaction using the data provided. SO82 (aq) + 3 l'(aq) 2 SO42-() + 13 "(aq) [S,08?) (M) (0) (M) Initial Rate (M's) 0.30 0.42 4.54 0.44 0.42 6.65 0.44 0.21 3.33 O Rate = 86 M?s" [S,08? Rate = 36 M 's' (S203 2701 Rate = 120 M2 [S208 2 101 Rate - 23 M-1/2" [S2O3 1912 Rate = 195 Mºss,Og 2012 Submit Request Answer