Question

Calculate $\Delta H$ in $kJ$ for the following reaction:
$C\left(g\right)+O_2\left(g\right)\to C{O}_2\left(g\right)$
Given that,

$H_{2}O\left(g\right)+C\left(g\right)\to CO\left(g\right)+H_2\left(g\right)$; $\Delta H=+131kJ$
$CO\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to {CO}_2\left(g\right)$; $\Delta H=-282kJ$

$H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to H_{2}O\left(g\right)$; $\Delta H=-242kJ$

• A. $-393$
• B. $+393$
• C. $+655$
• D. $-655$

Answer 1

The correct option is A $-393$

Given,
$H_{2}O\left(g\right)+C\left(g\right)\to CO\left(g\right)+H_2\left(g\right)$
$\Delta H=131kJ$ ........(i)
$CO\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to {CO}_2\left(g\right)$
$\Delta H=-252kJ$ .........(ii)
$H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to H_{2}O\left(g\right)$
$\Delta H=-242kJ$ ..........(iii)
$C\left(g\right)+O_2\left(g\right)\to C{O}_2\left(g\right)$ $\Delta H=$? ........(iv)
On adding equations (i), (ii) and (iii), we get equation (iv)
$H_{2}O\left(g\right)+C\left(g\right)\to CO\left(g\right)+H_2\left(g\right)$
$\Delta H=+131kJ$
$CO\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to C{O}_2\left(g\right)$
$\Delta H=-282kJ$
$H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to H_{2}O\left(g\right)$
_____________________________________________
$Cg+O_{2}g\to C{O}_{2}g$
$\Delta H=(131-282-242)kJ$
$=-393kJ$