Question

state common ion effect on solubility. The value of kp for the reaction Co2 + C ‐ 2CO is 3.0 at 1000k . If initially Pco2 = 0.48 bar and Pco = 0 bar and pure graphite is present .Calculate the equilibrium partial pressure of Co and Co2 .​

Answer 1

Answer:

The partial pressure of CO and CO_2CO

2 is 0.665 and 0.1475 bar respectively.

Explanation:

The balanced equilibrium reaction is: CO_2(g)+C(s)\rightleftharpoons 2CO(g)CO

2 (g)+C(s)⇌2CO(g)

At t=0 0.48 0 0

At t=t_{eq}t=t

eq

(0.48 - p) 2p

The expression used for K_pK

p

is,

K_p=\frac{(p_{CO})^2}{p_{CO_2}}K

p =

p

CO

2

(p co)

2

C does not come in this expression because it is in solid state and does not bear any pressure.

3=\frac{(2p)^2}{0.48-p}3=

0.48−p

(2p)

2

p = 0.3325 bar

Now we have to calculate the partial pressure of CO and CO_2CO

2

p_{CO}=2p=2\times 0.3325 bar=0.665 barp

CC

=2p=2×0.3325bar=0.665bar

p_{CO_2}=0.48- p=0.48-0.3325 bar=0.1475 barp

CO

2 =0.48−p=0.48−0.3325bar=0.1475bar

Therefore, the partial pressure of CO and CO_2CO

2 is 0.665 and 0.1475 bar respectively.

Explanation:

hope! it's helps