Question

Given that ΔH = −571.6 kJ/mol for the reaction 2 H2(g) + O2(g) → 2 H2O(l), calculate ΔH for these reactions. (a) 2 H2O(l) → 2 H2(g) + O2(g)

Answer 1

Answer : The value of $\Delta H$ for the reaction is +571.6 kJ/mole.

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 given chemical reaction is,

$2H_2(g)+O_2(g)\rightarrow 2H_2O(l)$     $\Delta H_1=-571.6kJ/mole$

Now we have to determine the value of $\Delta H$ for the following reaction i.e,

$2H_2O(l)\rightarrow 2H_2(g)+O_2(g)$    $\Delta H_2=?$

According to the Hess’s law, if we reverse the reaction then the sign of $\Delta H$ change.

So, the value $\Delta H_2$ for the reaction will be:

$\Delta H_2=-(-571.6kJ/mole)$

$\Delta H_2=+571.6kJ/mole$

Hence, the value of $\Delta H$ for the reaction is +571.6 kJ/mole.