Question

Use Hess's law and the following equations to calculate the ΔHreaction for the reaction CO(g) + 3H2(g) CH4(g) + H2O(g). (Show your work.) (4 points) • C(s) + O2(g) CO(g) ΔH = –110.5 kJ • C(s) + 2H2(g) CH4(g) ΔH = –74.85 kJ • H2(g) + O2(g) H2O(g) ΔH = –241.83 kJ

Answer 1

Answer: The enthalpy of reaction is, -206.18 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 final reaction is

$CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)$    $\Delta H_{rxn}=?$

The intermediate balanced chemical reaction will be,

(1) $C(s)+\frac{1}{2}O_2(g)\rightarrow CO(g)$     $\Delta H_1=-110.5kJ$

(2) $C(s)+2H_2(g)\rightarrow CH_4(g)$    $\Delta H_2=-74.85kJ$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(g)$    $\Delta H_3=-241.83kJ$

Now adding (2) and (3) and subtracting (1) , we get :

$\Delta H_{rxn}=(\Delta H_2+\Delta H_3)-\Delta H_1$

$\Delta H_{rxn}=((-74.85)+(-241.83))-(-110.5)$

$\Delta H_{rxn}=-206.18kJ$

Therefore, the enthalpy of reaction is, -206.18 kJ