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) leftharpoons 2CO(g) Kc for this reaction at the above temperature is

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) leftharpoons 2CO(g) Kc for this reaction at the above temperature is (A) 1.53 (B) 0.153 (C) 0.53 (D) 0.76

Tardigrade Admin · Accepted Answer

Let the total mass of the mixture of CO and CO2 is 100 g, then CO = 90.55 g and CO2 = 100 - 90.55 = 9.45 g Moles of CO=(90.55/28)=3.234 ; Moles of CO2=(9.45/44)=0.215 Mole fraction of CO =(3.234/3.234+0.215) =0.938 Mole fraction of CO2=(0.215/3.234+0.215) =0.062 ∴ pCO= mole fraction × total pressure =0.938×1 atm=0.938 atm pCO2=0.062×1 atm=0.062 atm Kp for the reaction, C(s) +CO2(g) leftharpoons 2CO(g) Kp=(p2CO/pCO2)=((0.938)2/0.062)=14.19 Now, Δ ng=2-1=1, Kp=Kc(RT)Δ ng or Kc=(Kp/RT)=(14.19/0.0821×1127)=0.153

Q. At $1127 K$ and $1 a t m$ pressure, a gaseous mixture of $CO$ and $C O_{2}$ in equilibrium with solid carbon has $90.55% CO$ by mass,
$C_{(s)} + C O_{2 (g)} ⇌ 2 C O_{(g)}$
$K_{c}$ for this reaction at the above temperature is

$1.53$

$0.153$

$0.53$

$0.76$

Q. At 1127K and 1atm 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)​ Kc​ for this reaction at the above temperature is

1.53

0.153

0.53

0.76

Q. At $1127 K$ and $1 a t m$ pressure, a gaseous mixture of $CO$ and $C O_{2}$ in equilibrium with solid carbon has $90.55% CO$ by mass,
$C_{(s)} + C O_{2 (g)} ⇌ 2 C O_{(g)}$
$K_{c}$ for this reaction at the above temperature is

$1.53$

$0.153$

$0.53$

$0.76$