(20pts.) The decomposition of ethylamine was followed at different temperatures.

Question

(20pts.) The decomposition of ethylamine was followed at different temperatures.

C2H5NH2 ==> C2H4 + NH3

The values of the total pressure as a function of time at three different temperatures are provided in the Excel file. The worksheet is setup for your convenience, but you do not have to use it. Each cell with a red triangle in the top right corner has information for you, just hover over the cell with your mouse to see it.

(Hint: First, calculate the partial pressure of the reactant at each point using the ICE box. Then calculate the rate constant for each temperature using the appropriate integrated rate equation. Finally, use the Arrhenius equation for the activation energy and pre-exponential coefficient).

Determine:

A. Order of the reaction

B. Rate constants at different temperatures

C. Energy of activation

D. Pre-exponential coefficient

T (K) 1/T (K 500 520 540 p (mmHg) Pc2SNH2 (mmH 120 360 600 64 79 89 72.5 92.5 103 84 104 slope: In k |slope: intercept E, (J/mol) NH3

Explanation / Answer

C2H5NH2 -----> C2H4 + NH3

After time 't' , PC2H5NH2 = (55-x) mm ; PC2H4 = PNH3 = x ml

at any time 't' total pressure of the gas mixture = (55+ x) mm

The reaction is a 1st order reaction

The governing equtions are -

rate constant(k)*time(t) = ln{[A0]/[A]} ; where [A0] = initial pressure of C2H5NH2 and [A] = pressure of C2H5NH2 after time 't'

thus, for the time interval t = 0 to t = 120 and temperature 500 0C

total pressure = 55+x = 64

or, x = 9 mm of Hg

Thus, k*120 = ln(55/46)

or, rate constant , k = 0.0015 s-1 .......(1)

for the time interval t = 0 to t = 120 and temperature 520 0C

total pressure = 55+x = 72.5

or, x = 17.5 mm of Hg

Thus, k*120 = ln(55/37.5)

or, rate constant , k = 0.0032 s-1 .........(2)

Now, as per Arrhenius equation ;-

lnk = lnA - (Ea/RT) ; where A = Arrhenius constant , Ea = energy of activation , R = universal gas constant, T = temperature in kelvin

Also, ln(k2/k1) = (Ea/R)*[(1/T1) - (1/T2)]

or, ln(0.0032/0.0015) = (Ea/8.314)*(20/773*793)

or, Ea = 193.073 KJ

Now, ln0.0032 = lnA - (193073/8.314*793)

or, lnA = 23.54

or, A = 1.64*1010

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