Question

Calculate Ho298 for the process Zn + S + 2O2 ? ZnSO4 from the following information:

Answer 1

Answer:  $\Delta H^0=-982.8kJ$.

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.

$Zn+S\rightarrow ZnS$    $\Delta H^0_1=-206.0kJ$   (1)

$ZnS+2O_2\rightarrow ZnSO_4$ $\Delta H^0_2=-776.8kJ$  (2)

The final reaction is:

$Zn+S+2O_2\rightarrow ZnSO_4$  $\Delta H^0_3=?$   (3)

By adding (1) and (2)

$\Delta H^0_3=\Delta H^0_1+\Delta H^0_2=-206.0kJ+(-776.8kJ)=-982.8kJ$

Hence $\Delta H^0=-982.8kJ$.