The following reaction: NO2 (g) --> NO (g) 1/2 O2 (g) is second-order in the reactant. The rate constant for this reaction is 3.40 L/mol*min. Determine the time needed for the concentration of NO2 to decrease from 2.00 M to 1.50 M.

t = 0.049 mins or 2.94 secs

Explanation:

For a simple second order reaction, the integrated law which describes the concentration of reactants at a given time t, is as follows: 1/[A] = 1/[A]o + Kt;

Where [A] is concentration of reactant at time, t, [A]o is initial concentration of A; K is rate constant; t is time at a given instant.

Using the integrated rate law:

I/[NO2]t - 1/[NO2]o = Kt

Where K = 3.40 L/mol/min

[NO2]t = 1.5 mol/L

[N02]o = 2.0 mol/L

t = ?

Making t subject of formula;

t = (1/[NO2]t - 1/[NO2]o) / K

t = (1/1.5 - 1/2.0)/3.40

t = 0.049 mins or 2.94 secs

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"