1 ) Determining a rate law from initial rates. lnitial rate data for the decompo
ID: 925114 • Letter: 1
Question
1 ) Determining a rate law from initial rates. lnitial rate data for the decomposition gaseous N2O5 at 55 degrees celsius are as follows: for experiment 1 initial N2O5 concentration is 0.020 M and the initial rate of decomposition of N2O5 (M/s) is 3.4 x 10^-5,for experiment 2 initial N2O5 concentration is 0.050 M and the initial rate of decomposition of N2O5 (M/s) 8.5 x 10^-5. A) What is the rate law? B) What is the value of the rate constant?C)What is the intial rate of decomposition of N2O5 at 55 degrees celsius when its initial concentration is 0.030 M?
2) 1. Rate constants for the gas-phase decomposition of hydrogen iodide, 2HI (g) H2 (g) + I2 (g), are listed in the following table:
Temperature (DEGREE CELSIUS) k (M^-1 s^ -1 )
283 3.52 x 10^-7
356 3.02 x 10^-5
393 2.19 x 10^-4
427 1.16 x 10-3
508 3.95 x 10^-2
a) Find the activation energy (in KJ/mol) using all 5 data points.
b) Calculate Ea from the rate constants at 283 C and 508 Degree C.
c) Given the rate constant at 283 degree Celsius and the value of Ea obtained in part (b), what is rate constant at 293 degree celsius?
Explanation / Answer
1)
A) Let rate law be represented as,
rate = k[N2O5]^x
with k = rate constant
then,
from experiment 1 and 2,
rate2/rate1 = 8.5 x 10^-5/3.4 x 10^-5 = (0.05/0.02)^x
taking log on both sides,
0.39794 = 0.39794^x
x = 1
So the rate law becomes,
rate = k[N2O5]^1
B) rate constant = rate/[N2O5] = 3.4 x 10^-5/0.02 = 1.7 x 10^-3 s-1
C) when initial concentration is 0.03 M
rate = 1.7 x 10^-3 x 0.03 = 5.1 x 10^-5 M/s
2)
a) Plot, 1/T on x-axis and lnK on y-axis
slope = Ea/R = 2.691
Ea = 2.691 x 8.314 = 22.373 J = 0.022 kJ/mol
b) Ea from rate cosnatnts at 283 and 508 oC
ln(k2/k1) = Ea/R[1/T1-1/T2]
ln(3.95 x 10^-2/3.52 x 10^-7) = Ea/8.314[1/556-1/781]
Ea = 186.58 kJ/mol
c) rate constant at 293 oC
ln(k2/3.52 x 10^-7) = 186.58/8.314[1/556 - 1/566]
rate cosnant at 293 oC = 3.5225 x 10^-7 M/s
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