Question

At 1000 K, Kp=19.9 for the reaction Fe2O3(s)+3CO(g)<--->2Fe(s)+3CO2(g) What are the equilibrium partial pressures of CO and if CO2 is the only gas present initially, at a partial pressure of 0.902 ?

Answer 1

Answer: The equilibrium concentration of CO is 0.243 atm

Explanation:

We are given:

Initial partial pressure of carbon dioxide = 0.902 atm

As, carbon dioxide is present initially. This means that the reaction is proceeding backwards.

For the given chemical equation:

                      $Fe_2O_3(s)+3CO(g)\rightleftharpoons 2Fe(s)+3CO_2(g)$

Initial:                                                                  0.902

At eqllm:                            3x                           (0.902-3x)

The expression of $K_p$ for above equation follows:

$K_p=\frac{(p_{CO_2})^3}{(p_{CO})^3}$

We are given:

$K_p=19.9$

Putting values in above equation, we get:

$19.9=\frac{(0.902-3x)^3}{(3x)^3}\\\\x=0.0810$

So, equilibrium concentration of CO = 3x = (3 × 0.0810) = 0.243atm[/tex]

Hence, the equilibrium concentration of CO is 0.243 atm

Answer 2

Answer:

0.243 bar; 0.659 bar  

Explanation:

The balanced equation is

Fe₂O₃(s) + 3CO(g) ⇌ 2Fe(s) + 3CO₂(g)

Data:

      Kp = 19.9

p(CO₂) =   0.902 bar

(You don't give the units for the partial pressure of CO₂, so I assume it is in bars}

1. Set up an ICE table.

$\begin{array}{ccccccc}\rm \text{Fe _{2} O}_{3}& + & \text{3CO} & \, \rightleftharpoons \, & \text{3Fe} & + & \text{3CO}_{2} \\ & & 0 & & & &0.902 \\ & & +3x & & & &-3x \\ & & 3x & & & &0.902 - 3x \\\end{array}$

2. Calculate the equilibrium concentrations

Initially, only CO₂ is present, so the reaction will go in the reverse direction and form CO.

$K_{\text{p}} = \dfrac{p_{\text{CO}_{2}}^{3}} {\text{p}_\text{CO}^{3}} = \dfrac{(0.902 - 3x)^{3}}{(3x)^{3}} = 19.9\\\\\begin{array}{rcl}\dfrac{0.902 - 3x}{3x} &=& 2.710\\\\0.902 - 3x = & = & 8.130x\\x & = & 0.0810\\ \end{array}$

p(CO) = 3x              = 3 × 0.0810       = 0.243 bar

p(CO₂) = 0.902 - 3x = 0.902 - 0.243 = 0.659 bar

Check:

$\begin{array}{rcl}\dfrac{0.659^{3}}{0.243^{3}}& =& 19.9\\\\\dfrac{0.286}{0.0143} & = & 19.9\\\\20.0 & = & 19.9\\\end{array}$

Close enough.