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

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2