Question

If $C(s) +O_{2}(g) \to CO_{2}(g), \Delta H= -X$
$CO(g) +\frac{1}{2} O_{2}(g) \to CO_{2}(g), \Delta H = -Y$
Calculate $\Delta_{t}H$ for $CO_{(g)}$ formation

• A. $- Y - X$
• B. $Y - X$
• C. $X + Y$
• D. $X - Y$

Answer 1

Answer: (B)

$C (S)+ O _{2}(g) \to CO _{2}(g) ; \Delta H_{1}=-x \quad \ldots( i )$

$CO (g)+\frac{1}{2} O _{2}(g) \to CO _{2}(g) ; \Delta H_{2}=-y \ldots(ii)$

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

$C(s)+O_2(g) \to CO_{2}(g)$
$\underset{(-)}{CO(g)}+\frac{1}{2} \underset{(-)}{O_{2}(g)} \to \underset{(-)}{CO_{2}(g)}$
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$C(s) +\frac{1}{2}O_{2}(g) \to CO(g)$
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$\therefore \Delta_{r}H$ for formation of $CO=\Delta\,H_{1}-\Delta\,H_{2}$

$=-x+y$ or $y-x$