Question

The N2O4−NO2 reversible reaction is found to have the following equilibrium partial pressures at 100∘C. Calculate Kp for the reaction. N2O4(g)⇌2NO2(g) 0.0005 atm 0.095 atm Express your answer using two significant figures.

Answer 1

Answer:

$K_{p}$ for the reaction is 18.05

Explanation:

Equilibrium constant in terms of partial pressure ($K_{p}$) for this reaction can be written as-

                $K_{p}=\frac{P_{NO_{2}}^{2}}{P_{N_{2}O_{4}}}$

where $P_{NO_{2}}$ and $P_{N_{2}O_{4}}$ are equilibrium partial pressure of $NO_{2}$ and $N_{2}O_{4}$ respectively

Hence $K_{p}=\frac{(0.095)^{2}}{(0.0005)}$ = 18.05

So, $K_{p}$ for the reaction is 18.05