Question

The half-life for the first-order of decomposition is 1.3 x 10^-5s N2O4(g) 2N02(g) If N2O4 is introduced into an evacuated flask at a rate of 17.0mm Hg, how many seconds are required for the No2 to reach 1.3 mm Hg? To solve this problem one plots a graph of______ versus _____. the slope of this lie is equal to negative activation energy divided by ______and has the value ______ 104 K. (2 s.f.) The activation energy is ______ kJ/mole (2 s.f.) Fill in the blanks.

Answer 1

t-half of first order reaction

t1/2 = 0.693/k

k = 53307.69

N2O4 ---- > 2N02

for ideal gas, PV=nRT

so the total number of moles will be same intially and finally.

and moles of NO2 is two times that of N2O4

at time of interest pNO2 = 1.3 mmHg = 1.3/17 times p N2O4 initial = 0.076

so pNO2 reacted will be 1/2 of 0.076 = 0.038

begining amount = 1, end amount is (1-0.038) = 0.962

now

B/Bi = e^(-kt)

t = time elapsed.

So now we can use it to calculate the Activation Energy by graphing lnk versus 1/T.

ANswer to your fill in blanks is 9(lnK) vs 1/T, Slope is equal to negative of Ea devided by R(gas constant) = 8.314 J/mol-K

so Ea/R = 104

so Ea = 104*8.314 = 864.65 kJ /mol

Answer. Hope this helps you clearing your doubts.