Question

Gaseous carbon dioxide is partially decomposed according to the following equation. An initial pressure of 1.00 atm of CO2 is placed in a closed container at 2500 K, and 2.1 % of the molecules decompose. Determine the equilibrium constant Kp at this temperature

Answer 1

Answer:

Kp = 24.8

Explanation:

$2CO_{2(g)}$ ⇄ $2CO + 0_{(2g)}$

1.00            0        0

1.0-2x        2x       x

x = 2.1%

$= 1 X \frac{2.1}{100}$

= 0.021

$P_{co_{2}} = 1- 2 X 0.021$

=0.958 atm

$P_{co = 0.042 atm$

$P_{o_{2}} = 0.021 atm$

$Kp = \frac{[CO_{2} ]^{2} }{[CO]^{2} [O_{2} ] }$

$Kp = \frac{[0.958 ]^{2} }{[0.042]^{2} [0.021] }$

=$\frac{0.917}{(1.76 X 10^{-3} ) (0.021 )}$

=0.917/0.0000369

=24.8