Question

Dinitrogentetraoxide partially decomposes into nitrogen dioxide. A 1.00-L flask is charged with 0.0400 mol of N2O4. At equilibrium at 373 K, 0.0055 mol of N2O4 remains. Keq for this reaction is ________.

Answer 1

Keq=0.866

Step-by-step explanation:

Hello,

In this case, the undergone chemical reaction is:

`N_2O_4<->2NO_2`

In such a way, since 0.0055 mol of N₂O₄ remains in the flask, one infers that the reacted amount (
`x`) was:

`x=0.04mol-0.0055mol=0.0345mol`

In addition, the produced amount of NO₂ is:

`2*0.0345mol=0.069mol`

Finally, considering the flask's volume, the equilibrium constant is then computed as follows:

`Keq=((2*0.0345M)^2)/(0.0055M)=0.866`

Best regards.