Question

The following equilibrium constants were determined at $1120\, K :$
$2 CO ( g ) \rightleftharpoons C ( s )+ CO _{2}( g ) ; K_{P_{1}}=10^{-14} atm ^{-1}$
$CO ( g )+ Cl _{2}( g ) \rightleftharpoons COCl _{2}( g ) ; K_{P_{2}}=6 \times 10^{-3} atm ^{-1}$
What is the equilibrium constant $K_{C}$ for the following reaction at $1120\, K$ :
$C ( s )+ CO _{2}( g )+2 Cl _{2}( g ) \rightleftharpoons 2 COCl _{2}( g )$

• A. $3.31 \times 10^{11} M ^{-1}$
• B. $5.5 \times 10^{10} M ^{-1}$
• C. $5.51 \times 10^{6} M ^{-1}$
• D. None of these

Answer 1

Answer: (A)

$C ( s )+ CO _{2}( g ) \rightleftharpoons 2 CO ( g );$...(1)

$K_{P_{1}}=10^{14} atm ^{-1}$

$2 CO ( g )+2 Cl _{2}( g ) \rightleftharpoons 2 COCl _{2}( g );$...(2)

$K_{P_{2}}=\left(6 \times 10^{-3}\right)^{2} atm ^{-2}$

Add (1) and (2), we get

$C( s )+ CO _{2}( g )+2 Cl _{2}( g ) \rightleftharpoons 2 COCl _{2}( g );$

$K_{P}=10^{14} \times 36 \times 10^{-6}=36 \times 10^{8}$

For given reaction $n_{ g }=-1$

$\therefore K_{C}=K_{P}(R T)=36 \times 10^{8} \times 0.0821 \times 1120$

$K_{C}=3.31 \times 10^{11} M ^{-1}$