Question

The half-life for the first-order decomposition of N204is 1.3 x 10-5s

N2O4 (g)    arrow sign      2NO2(g)

If N204is introduced into an evacuated flask at a pressure of 17.0 mm Hg, how many seconds are required for the pressure of NO2 to reach 1.3 mm Hg?

What is the value of the rate constant?_____/second (2 s.f.) What is the elapsed time for the reaction? _____seconds (2 s.f.) use E to express powers of ten, i.e. 6.02x1023 =6.02E23

Please fill in the blanks to help me understand clearly the answer or put a box around it. Thank you.

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