Rate Law from Concentration versus Time Data The rate of the above reaction depe
ID: 999884 • Letter: R
Question
Rate Law from Concentration versus Time Data
The rate of the above reaction depends only on the concentration of nitrogen dioxide at temperatures below 225°C.
At a temperature below 225°C the following data were obtained:
Time
(s)
[NO2]
(mol/L)
0
0.800
1.85×103
0.669
4.62×103
0.537
6.94×103
0.462
1.39×104
0.325
2.78×104
0.204
(1) Which of the following expressions for the rate law (either differential or integrated) are completely consistent with the above experimental data.
Yes No 1/[NO2] = kt + 1/[NO2]0
Yes No ln([NO2]/[NO2]0) = -kt
Yes No d[NO]/dt = k[NO2]2
Yes No k[NO2]2 = -d[CO]/dt
Yes No k[NO2] = -d[NO2]/dt
(2) Calculate the rate constant (k). Enter the numerical value and corresponding units.
NO2(g) + CO(g) NO(g) + CO2(g)Explanation / Answer
!) rate law : ln ([NO2]/[NO2]0) = -kt i.e first order
2) plot of ln (NO2) vs t (neglecting the last two values) gives the following equation:
y = -78.882x-0.2403
here slope = k = 78.882 s-1
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