Discussion Forum

1)

If $C(s)+O_{2}(g)\rightarrow CO_{2}(g);$  $\triangle H_{}=-X$     

$CO(g)+\frac{1}{2}O_{2}(g)\rightarrow CO_{2}(g);$   $\triangle H_{}=-Y$  .

Calculate  $\triangle _{f}H$ for CO (g) formation

Answer:

Option B

Explanation:

$C(s)+O_{2}(g)\rightarrow CO_{2}(g);\triangle H_{1}=-x$     ...........(i)

$CO(g)+\frac{1}{2}O_{2}(g)\rightarrow CO_{2}(g);\triangle H_{2}=-y$  ...........(ii)

 For the formation of CO subtract Eqs.(ii) from (i)  i.e,

  $C(s)+O_{2}(g)\rightarrow CO_{2}(g)$

$\underline{CO(g)_{(-)}+\frac{1}{2}O_{2}(g)_{(-)}\rightarrow CO_{2}(g)_{(-)}}$

$\underline{C(s)+\frac{1}{2}O_{2}(g)\rightarrow CO(g)}$

 $\therefore$    $\triangle_{r}H$ for formation of CO=  $\triangle H _{1}-\triangle H _{2}$

                                                                    =-x+y   or y-x