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# At 1127 K and 1 Atm Pressure, a Gaseous Mixture of Co and Co2 In Equilibrium with Solid Carbon Has 90.55% Co by Mass Calculate Kc for this Reaction at the Above Temperature. - Chemistry

#### Question

At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with solid carbon has 90.55% CO by mass

C (s) + CO2 (g) ⇌ 2CO (g)

Calculate Kc 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 CO2 = (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 CO2,

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= 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`

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