Question

Determine $\Delta H$ for the following reaction at 500K and constant pressure :
$CO\left(g\right)+H_{2}O\left(g\right)\to C{O}_2\left(g\right)+H_2\left(g\right)$
use the following data :

Substance	$C_P$ ( J / mol K )	${\Delta }_{f}H$ ( 298K ) ( kJ / mol )
CO	29.12	$-$110.5
$H_{2}O$	33.58	$-$241.8
$C{O}_2$	37.11	$-$393.5
$H_2$	29.89	0.0

• A. $\Delta H=-30.3$ kJ
• B. $\Delta H=-50.3$ kJ
• C. $\Delta H=-40.3$ kJ
• D. $\Delta H=-20.3$ kJ

Answer 1

The correct option is C $\Delta H=-40.3$ kJ

At two different temperatures ($T_1$=298K and $T_2$=500K),
the corresponding enthalpy changes are related as : $H_2$ - $H_1$ = ($T_1$ - $T_1$) [$C_p$(products) - $C_1$(reactants)]..... (1)
Now, $H_1$ = [${\Delta }_f{H}_{{CO}_2}$(g) + ${\Delta }_f{H}_{H_2}$(g)] - [${\Delta }_f{H}_{CO}$(g) + ${\Delta }_f{H}_{H_{2}O}$(g)],
$H_1$= -393.5 + 241.8 + 110.5 = -41.2 kJ/mol,
$C_p$(products) = 29.89 + 37.11= 67 J/molK, $C_p$(reactants) = 29.12 + 33.58 J/molK,
Therefore, plugging these values in equation (1), we get: $H_2$ - 41.2 = $\frac{(500-298)(67-62.7)}{1000}$, $H_2$ = -40.3 kJ/mol at 500 K