#### Question

At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO_{2} in equilibrium with solid carbon has 90.55% CO by mass

C (s) + CO_{2} (g) ⇌ 2CO (g)

Calculate K_{c} for this reaction at the above temperature.

#### Solution 1

Let the total mass of the gaseous mixture be 100 g.

Mass of CO = 90.55 g`

And, mass of CO_{2} = (100 – 90.55) = 9.45 g

Now, number of moles of CO, `n_(CO) = 90.55/28 = 3.234`

Number of moles of `CO_2, n_(CO_2)` = 9.45/44 = 0.215 mol

Partial pressure of CO,

`p_(CO) = n_(CO)/(n_(CO) + n_(CO_2))xx p_"total"`

`= 3.234/(3.234 + 0.215) xx 1`

= 0.938 atm

Partial pressure of CO_{2},

`p_(CO_2) = n_(CO_2)/(n_(CO) + n_(CO_2)) xx p_"total"`

`=0.215/(3.234 + 0.215) xx 1`

Therefore `K_p = [CO]^2/[CO_2]`

`= (0.938)^2/0.062`

= 14.19

For the given reaction,

Δ*n *= 2 – 1 = 1

We know that,

`K_P = K_C(RT)^(trianglen)`

`=> 14.19 = K_C(0.082 xx 1127)^1`

= 0.154 (approximately)

#### Solution 2

Step 1: Calculation of `K_p` for the reaction

Let the total mass of gaseous mixture = 100 g

Mass of CO in the mixture = 90.55 g

Mass of `CO_2` in the mixture = (100 - 90.55) = 9.45 g

No of moles of CO = `90.55g/((28g mol^(-1))) = 3.234 mol`

No of moles of `CO_2` = `9.45/(44 g mol^(-1))` = 0.215 mol

`P_(CO)` in the mixture = `(3.234 mol)/(3.234 + 0.215 ) xx 1 atm = (3.234 mol)/(3.499) xx 1 atm = 0.938 atm`

`p_(CO_2)` in mixture = `(0.215 mol)/(3.449 mol) xx 1 atm = 0.062 atm`

`C(s) + CO_2(g) ⇌ 2CO(g)`

Eqm pressure 0.062 atm 0.938 atm

`K_p = (p^2CO)/(pCO_2) = (0.938 atm)^2/(0.062 atm) = 14.19 atm`

Step II. Calculation of `K_C` for reaction

`K_C = (K_p)/(RT)^(triangleng)`

`K_p = 14.19 "atm", R = 0.0821 "L atmK"^(-1) mol^(-1), T = 1127 K triangle^(ng) = 2 - 1 = 1`

`K_c = (14.19 atm)/((0.0821L atm K^(-1) mol^(-1)) xx (1127 K)^1) = 6.46`