From the following data
(i) `H_(2)(g)+CO_(2) (g) Leftrightarrow H_(2)O (g)+CO(g), K_(2000)K=4.40`
(ii) `2H_(2)O (g) Leftrightarrow 2H_(2) (g) +O_(2) (g), K_(1000 K)=2.24 xx 10^(22)`
show whether the reaction (iii) is exothermic or endothermic.

Explanation

Let us consider reactions (i) and (ii).
Multiplying Equation (i) by 2, we get,
(iv) `2H_(2) (g)+2CO_(2) (g) Leftrightarrow 2H_(2)O (g)+2CO(g)`
`K_(2000 K)^(II) =K_(2000 K)^(2) =4.40^(2) ……..("Eqn 6")`
Adding (iv) and (ii), we have
`2CO_(2) (g) Leftrightarrow 2CO+O_2`
`K_(2000 K)^(III)=K_(2000 K)^(II) xx K_(2000 K)^(I) .......(Eqn 7)`
by reversing the reaction, we get the reaction (iii) but at 2000 K
(v) `2CO (g)+O_2 (g) Leftrightarrow 2CO_2 (g)`, for which,
`K_(2000 K)^(IV)=(1)/(K_(2000 K)^(III) ...("Eqn .5")`
`=(1)/(4.40^(2) xx 5.31 xx 10^(-10))`
`=9.72 xx 10^(7)`
From (iii) and (v) we see that as the temperature increases, the equilibrium constant decreases `(9.72 xx 10^(7) lt 2.24 xx 10^(22))`. The reaction (iii) is thus exothermic because `triangleH^(0)` will be negative.