Using The Following Data, Determine the Rate Law For The Reaction.
Part A.
2NO(g)+O(g) --> 2NO2(g)
Experiment |
[NO] |
[O2] |
Initial Rate (M/s] |
1 |
0.0126 |
0.0125 |
1.41x10^-2 |
2 |
0.0252 |
0.0250 |
1.13x10^-1 |
3 |
0.0252 |
0.0125 |
5.64x10^-2 |
Part B.
S2O8charge2-(ag)+3 I charge 1-(aq) ----> 2SO4charge2-(aq)+I3charge1-(aq)
Experiment |
[S2O8charge2-] |
[ I charge1-] |
Initial Rate (M/s) |
1 |
0.023 |
0.048 |
6.8x10^-6 |
2 |
0.054 |
0.048 |
1.6x10^-5 |
3 |
0.054 |
0.019 |
6.3x10^-6 |
Part A
from the table one while moving from first column to last column
if concentration No doubles rate increased two times that means wrt to NO second order
simlarly while moving from third column to second column concentration od O2 doubles rate become doubled so wrt O2 it is first order
rate law
rate = k [NO]2 [O2]1
second part is little confusing plaz ask somebady
Using The Following Data, Determine the Rate Law For The Reaction. Part A. 2NO(g)+O(g) --> 2NO2(g)...
The following data were collected for the rate of disappearance ofNO in the reaction 2NO(g) + O2 --> 2NO2(g) : Experiment [NO] (M) [O2] (M) Initial Rate (M/s) 1 0.0126 0.0125 1.41 x 10-2 2 0.0252 0.0125 5.64 x 10-2 3 0.0252 0.0250 1.13 x 10 -1 (a) What is the rate law for the reaciton? (B) What are teh unitsof the rate constant? (c) What is the average value of the rateconstant calculated form teh three data sets? (d)...
Consider the following reaction: 2NO(g)+O2(g)⇌2NO2(g)2NO(g)+O2(g)⇌2NO2(g) The data in the table show the equilibrium constant for this reaction measured at several different temperatures. Temperature K 400 1.9x10^7 425 2.5x10^6 465 1.6x10^5 515 8.8x10^3 600 2.0x10^2 Use the data to find ΔrS∘ΔrS∘ for the reaction. (answer in J K-1 mol-1) The answer is NOT 158
4. . The reaction 2NO(g) + O2(g) → 2NO2(g) has the following rate law: Rate = k[O2][NO]2 . If the concentration of NO is reduced by a factor of two, the rate will __________
Please show all work Determine the rate law for the reaction and calculate the rate constant Experiment [NO]. [O2). Initial rate ignore this |(M/s) column 1 0.0235 0.0125 7.98E-3 2 0.0235 0.0250 15.9E-3 3 0.0470 0.0125 32.0E-3 4 0.0470 0.0250 63.5E-3
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]
The reaction 2NO(g) + O2(g) + 2NO2(g) was studied, and the following data were obtained where A[02] Rate At [NO). [02] Initial Rate (molecules/cm) (molecules/cm) (molecules/cms) 1.00 x 1018 3.00 x 1018 2.50 x 1018 1.00 x 1018 1.00 x 1018 2.50 x 1018 2.00 x 1016 1.80 x 1017 3.13 x 1017 What would be the initial rate for an experiment where [NO], = 1.24 x 1018 molecules/cm and [02], = 7.50 x 1018 molecules/cm'? molecules/cm. Rate = -S
Given the reaction and following data: 2NO + O2 → 2NO2 Calculate the rate law constant and write the rate law. Trial [NO] (mol/L) [O2] (mol/L) Rate (mol/L/hr) 1 0.125 0.125 2.57 x 10-2 2 0.250 0.125 5.20 x 10-2 3 0.250 0.250 4.16 x 10-1
Initial rate data is given in the table for the following reaction. 2NO(g) +0,(g) + 2NO,(8) Experiment (NO) initia, (M) olinitial (M) Rate (M/s) 0.420 0.460 0.206 2 0.840 0.460 0.824 3 / 0.420 0.920 0.412 4 0.420 0.230 0.103 What is the rate law for the reaction? Orate = k[NO][02] Orate = k[NO]”[02] rate = k[NO] rate = k[NO]{0,12
06 Question (3 points) At a given temperature, K 8.6x10 for the following forward reaction: 2NO(g)+02(8) 2NO2(g) Write the reverse reaction. Be sure to include all states of matter. x. x (aq). He 2NO (g) 2NO(g) +02(g) Part 2 (1.5 points) Feedback What is the Kvalue for the reaction above written in the reverse direction? 0,00000116
2. (10 pts) Consider the following second-order reaction: 2NO2(g) + 2NO(g)+O2(g) The rate constant is 0.54 M-15-1, at 300 °C. How long would it take for the concentration of NO2 to decrease from 0.62M to 0.28M?