Question

The standard enthalpy change for the following reaction is -971 kJ at 298 K What is the standard enthalpy change for the reaction at 298 K? 4 Mn(s) + 3。2(g)-→ 2 Mn2O3(s) kJ

Answer 1

Enthalpy is an extensive property i.e. it depends on the amount of reactants.

if the amount of reactants double, as in the above reaction, the enthalpy will also double. Therefore, standard enthalpy for

4Mn + 3O_{2 $\rightarrow$} 2Mn₂O₃ will be -1942kJ

Enthalpy is also a state function so the standard enthapy for the reaction

Be(s) + O₂(g) + H₂ (g) $\rightarrow$ Be(OH)₂ (s) will be -903kJ