Question

The value of Kp for the reaction NO(g) 1 1 2 O2(g) 4 NO2(g) is 1.5 3 106 at 25°C. At equilibrium, what is the ratio of PNO2 to PNO in air at 25°C? Assume that PO2 5 0.21 atm and does not change.

Answer 1

Answer : The ratio of $p_{NO_2}$ to $p_{NO}$ is, $6.87\times 10^5$

Solution :  Given,

$K_p=1.5\times 10^6$

$p_{O_2}$ = 0.21 atm

The given equilibrium reaction is,

$NO(g)+\frac{1}{2}O_2\rightleftharpoons NO_2(g)$

The expression of $K_p$ will be,

$K_p=\frac{(p_{NO_2})}{(p_{NO})\times (p_{O_2})^{\frac{1}{2}}}$

Now put all the given values in this expression, we get:

$1.5\times 10^6=\frac{(p_{NO_2})}{(p_{NO})\times (0.21)^{\frac{1}{2}}}$

$\frac{(p_{NO_2})}{(p_{NO})}=(1.5\times 10^6)\times (0.21)^{\frac{1}{2}}$

$\frac{(p_{NO_2})}{(p_{NO})}=6.87\times 10^5$

Therefore, the ratio of $p_{NO_2}$ to $p_{NO}$ is, $6.87\times 10^5$