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, $C(s)+CO_{2}(g) \rightleftharpoons 2CO(g)$, Calculate Kc for this reaction at the above temperature.

Answer 1

Let us consider the weight of mixture is 100 g,

In that, 90.55% CO by mass means 90.55 g of CO and 9.45 g of CO2 present in the mixture.

$Number\quad of\quad moles\quad of\quad CO\quad =\quad \frac { weight\quad of\quad CO }{ molar\quad mass\quad of\quad CO } \quad \left$

$number\quad of\quad moles\quad =\quad \frac { weight }{ molar\quad mass } \right)$

$= \frac { 90.55 }{ 28 } = 3.234 mol$ [ molar mass of CO = 28 g/mol ]  

$Number\quad of\quad moles\quad of\quad { CO }_{ 2 }\quad =\quad \frac { given\quad weight\quad of\quad { CO }_{ 2 } }{ molar\quad mass\quad of\quad { CO }_{ 2 } }$

$= \frac { 9.45 }{ 44 } = 0.215 mol$ [ molar mass of ${ CO }_{ 2 } = 44g/mol$ ]  

$Mole\quad Fraction\quad of\quad CO\quad =\quad \frac { nCO }{ nCO } \quad +\quad n{ CO }_{ 2 }$

$=\quad \frac { 3.234 }{ 3.234 } \quad +\quad 0.215\quad =\quad 0.938$

$Mole\quad Fraction\quad of\quad { CO }_{ 2 }\quad is\quad \left( 1\quad -\quad { x }_{ CO } \right) \quad =\quad 1\quad -\quad 0.938\quad =\quad 0.062$

Partial pressure of CO = mole fraction of CO $\times$  total pressure  

= 0.938  $\times$ 1 atm  

Partial pressure of CO = 0.938 atm

Similarly, partial pressure of ${ CO }_{ 2 } = mole fraction\quad of\quad { CO }_{ 2 } \times total\quad pressure$

= 0.062 × 1 atm

Partial pressure of ${ CO }_{ 2 }$ = 0.062 atm  

Then, ${ C }_{ (s) }\quad +\quad { CO }_{ 2(g) }\quad =\quad 2{ CO }_{ (g) }$

${ K }_{ p }\quad =\quad \frac { { \left( 0.938 \right) }^{ 2 } }{ 0.062 } \quad =\quad 14.19$

∆ng = number of mole of gaseous products - number of mole of gaseous reactants  

= 2 - 1 = 1  

${ K }_{ c }\quad =\quad \frac { { K }_{ p } }{ RT }$  

${ K }_{ c }\quad =\quad \frac { 14.19 }{ 0.082 } \quad \times\quad 1127\quad [T = 1127 K and\quad R = 0.0821 L.atm/K/mol]$

${ K }_{ c }\quad =\quad 0.15336\quad \approx \quad 0.153$