Question

Question 7.23 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.

Class XI Equilibrium Page 226

Answer 1

Let weight of mixture is 100g , then 90.55% CO by mass means 90.55g CO and 9.45g CO2 present in the mixture.

no of moles of CO = Given weight of CO/molar mass of CO
= 90.55/28 = 3.234 mol [ molar mass of CO = 28 g/mol ]

no of moles of CO2 = given weight of CO2/molar mass of CO2
= 9.45/44 = 0.215 mol [ molar mass of CO2 = 44g/mol ]

$mole \:fraction \:of\: CO(x_{CO})=\frac{n_{CO}}{n_{CO}+n_{CO_2}} \\ = \frac{3.234}{3.234 + 0.215} = 0.938 \\ \\ \\ mole \: fraction \: of \: CO_2 = (1 -x_{CO}) = 1 - 0.938 = 0.062$
now, partial pressure of CO = mole fraction of CO × total pressure
= 0.938 × 1 atm
= 0.938 atm

similarly, partial pressure of CO2 = mole fraction of CO2 × total pressure
= 0.062 × 1 atm
= 0.062 atm

now,
C(s) + CO2(g) = 2CO(g)
$K_p=\frac{P_{[CO]^2}}{P_{[CO_2]^1}}$
Kp = (0.938)²/(0.062) = 14.19
∆ng = number of mole of gaseous products- number of mole of gaseous reactants
= 2 - 1 = 1
$now, K_p =K_c(RT)^{\triangle{n_g}}$
Kc = Kp/RT
= 14.19/0.0821 × 1127 [ T = 1127 K and R = 0.0821 L.atm/K/mol ]
= 0.15336 ≈0.153