Question

A reaction mixture at 175 K initially contains 522 torr of NO and 421 torr of O2. At equilibrium, the total pressure in the reaction mixture is 748 torr. Calculate Kp at this temperature. Express your answer to three significant figures.

Answer 1

Answer:

$Kp=0.0386$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2NO+O_2\rightleftharpoons 2NO_2$

For which the equilibrium expression is:

$Kp=\frac{p_{NO_2}^2}{p_{NO}^2p_{O_2}}$

Whereas, at equilibrium, each pressure is computed in terms of the initial pressure and the reaction extent via:

$p_{NO_2}=2x\\p_{NO}=522-2x\\p_{O_2}=421-x$

And the total pressure:

$p_{eq}=p_{NO_2}+p_{NO}+p_{O_2}\\\\p_{eq}=2x+522-2x+421-x\\\\p_{eq}=943-x$

Yet it is 748 torr, for which the extent is:

$x=943-p_{eq}=943-748\\\\x=195torr$

Therefore, Kp turns out:

$Kp=\frac{(2x)^2}{(522-2x)^2(421-x)}\\\\Kp=\frac{(2*195)^2}{(522-2*195)^2(421-195)}\\\\Kp=0.0386$

Best regards.