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

Answer:

Keq=0.866

Explanation:

Hello,

In this case, the undergone chemical reaction is:

$N_2O_42NO_2$

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:

$Keq=\frac{(2*0.0345M)^2}{0.0055M}=0.866$

Best regards.