Question

At 450 K, Kp = 2.0 × 1010 bar for the given reaction at equilibrium. $2SO_{2}(g)+O_{2}(g) \rightleftharpoons 2SO_{3}(g)$, What is Kc at this temperature ?

Answer 1

$2{ SO }_{ 2(g) }\quad +\quad { O }_{ 2(g) }\quad \rightleftharpoons \quad 2{ SO }_{ 3(g) }$

Here, ∆ng = number of moles of gaseous products - number of moles of gaseous reactants  

= 2 - (1 + 2) = -1  

We know that, ${ K }_{ p }\quad =\quad 2\quad \times \quad { 10 }^{ 3 }\quad /bar$

R = 0.0831 L.bar/K/mol  

T = 450 K

${ K }_{ c }\quad =\quad \frac { { K }_{ p } }{ { RT }^{ (-1) } }$  [because , ∆ng = -1]  

$=\frac { { K }_{ p } }{ { RT }^{ (-1) } }$

$=\quad 2\quad \times \quad { 10 }^{ 10 }\quad \times \quad 0.0831\quad \times \quad 450$  

$=\quad 7.479\quad \times \quad { 10 }^{ 11 } L/mol$

At this temperature, ${ K }_{ c }\quad =\quad 7.479\quad \times \quad { 10 }^{ 11 } L/mol$