Question

2C(s)+O2(g)2H2(g)+O2(g)H2O(l)→2CO(g)→2H2O(g)→H2O(g)ΔHΔHΔH=−222kJ=−484kJ=+44kJ Use the thermochemical data above to calculate the change in enthalpy for the reaction below. H2O(l)+C(s)→CO(g)+H2(g)

Answer 1

Answer : The change in enthalpy for the reaction is, 175 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 main reaction is:

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

The intermediate balanced chemical reaction will be,

(1) $2C(s)+O_2(g)\rightarrow 2CO(g)$     $\Delta H_1=-222kJ$

(2) $2H_2(g)+O_2(g)\rightarrow 2H_2O(g)$    $\Delta H_2=-484kJ$

(3) $H_2O(l)\rightarrow H_2O(g)$    $\Delta H_3=44kJ$

Now we are dividing reaction 1 by 2, dividing reverse reaction 2 by 2 and then adding all the equations, we get :

(1) $C(s)+\frac{1}{2}O_2(g)\rightarrow CO(g)$     $\Delta H_1=\frac{-222kJ}{2}=-111kJ$

(2) $H_2O(g)\rightarrow H_2(g)+\frac{1}{2}O_2(g)$    $\Delta H_2=\frac{484kJ}{2}=242kJ$

(3) $H_2O(l)\rightarrow H_2O(g)$    $\Delta H_3=44kJ$

The expression for change in enthalpy of the given reaction is:

$\Delta H=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(-111)+(242)+(44)$

$\Delta H=175kJ$

Therefore, the change in enthalpy for the reaction is, 175 kJ