) Dinitrogen Tetroxide partially decomposes according to the following equilibrium: N2O4 (g) 2NO2 (g) A 1.00-L flask is charged with 0.400 mol of N2O4. At equilibrium at 373 K, 0.0055 mol of N2O4 remains. Keq for this reaction is .

The value of for this reaction is 1.578.

Explanation:

Given: Initial moles of = 0.4 mol

Volume = 1.00 L

Therefore, initial concentration of is calculated as follows.

Now, ICE table for the given reaction equation is as follows.

Initial:               0.4               0

Change:           -x                +2x

Equib:  0.4 - x = 0.0055       2x

Hence, the value of x is calculated as follows.

0.4 - x = 0.0055

x = 0.4 - 0.0055

= 0.3945

Now, the is calculated as follows.

2x = =

Therefore, for the given reaction is calculated as follows.

Thus, we can conclude that for this reaction is 1.578.

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

while cooking a piece of 1/2 pound beef steak, a cast iron pan (2.4kg in weight) was heated 260 celsius.