Question

QUESTION 13
Use the standard reaction enthalpies given below to determine AHºrn for the followi
C(s) + H2O(g) - CO(g) + H2(9)
Given:
Reaction 1: C(s) + O2(g) – CO2(g) AHºrxn = -605 kJ
Reaction 2: 2 CO(g) + O2(g) – 2 CO2(g) AH°x = -966 kJ
Reaction 3: 2 H2(g) + O2(g) → 2 H2O(g) AHⓇx = -638 kJ​

Answer 1

Answer: The $\Delta H^0{rxn}$ for the reaction is 212 kJ

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The net reaction is

$C(s)+H_2O(g)\rightarrow CO(g)+H_2(g)$    $\Delta H=?$

The intermediate balanced chemical reaction will be,

(1) $C(s)+O_2(g)\rightarrow CO_2(g)$     $\Delta H_1=-605kJ$

(2) $2CO(g)+O_2(g)\rightarrow 2CO_2(g)$    $\Delta H_2=-996kJ$

(3) $2H_2(g)+O_2(g)\rightarrow 2H_2O(g)$    $\Delta H_3=-638kJ$

$\Delta H=\Delta H_1+\frac{1}{2}\times -\Delta H_2+\frac{1}{2}\times -\Delta H_3$

$\Delta H=(-605)+\frac{1}{2}\times (996)+\frac{1}{2}\times (638)$

$\Delta H=212kJ$

Therefore, the $\Delta H^0{rxn}$ for the reaction is 212 kJ