2. The decomposition of N2O4 to form NO2 is a first order reaction. At 10.00 C,

ID: 483825 • Letter: 2

Question

2. The decomposition of N2O4 to form NO2 is a first order reaction.

At 10.00 C, the rate constant (k) value was determined to be 4.50 * 10 ^3 / sec

with an (Ea) activation energy of 58.0 KJ / mol

a. write a rate law for this reaction

b. In a graphical representation of the decomposition of N2S04 to NO2 what vaiable would be on the Y- axis?

c. Determine the temperature (K) of the reaction in Kelvin when the rate constant's (K) value is 1.0 * 10^4/s

d. Calculate the half life for this reaction at 10.0 C

If first order, then it depends on N2O4 directly

RAte = -k*[N2O4]^1

typically, we choose ln([N2O2]) since this is firs torder, so we recommend compare natural log of concentaiton vs. time(inx-axis

Fin T for k = 10^4

from ahrrenius:

K = A*exp(-E/(RT))

from data, find A

A = K/(exp(-E/(RT)) = (4.5*10^3)/(exp(-58000/(8.314*283))

A = 2.28523*10^14

so..

K = A*exp(-E/(RT))

converts to

K = 2.28523*10^14*exp(-58000/(8.314*T))

choose K = 10^4 and solve for T

10^-4 = 2.28523*10^14*exp(-58000/(8.314*T))

-58000/(8.314*T) = ln( (10^-4)/ (2.28523*10^14))

T = -58000/(8.314*-42.272) = 165.03086 K

T = 165.03086-273 = -107.9691 °C

find half life at T = 10°C

t 1/2 = ln(2)/k for a 1st rate reaction

t 1/2 = ln(2)/(4.5*10^3) = 0.0001540 seconds

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