Question

The standard enthalpies of formation \[{\rm{A}}{{\rm{l}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}\] and CaO are -1675 \[{\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}\] and -635 \[{\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}\]respectively. For the reaction, \[3{\rm{CaO}} + {\rm{2Al}} \to {\rm{3Ca + A}}{{\rm{l}}_2}{{\rm{O}}_{\rm{3}}}\] , the standard enthalpy \[{\Delta _r}{H^0}\] =--kJ.. (Round off to the nearest integer)

Answer 1

Hint: For a chemical reaction, the stand enthalpy can be calculated by subtracting the enthalpy reactants from the enthalpy of products in their standard states. The enthalpy of a reactant or product present in the solid or gaseous state is zero.

Formula used: The formula used in the calculation of enthalpy is,
\[{\Delta _r}{H^0} = {\Delta _r}{H^0}_f{\rm{(Products}}) - {\Delta _r}{H^0}_f{\rm{(Reactants}})

Complete Step by Step Solution:
The given reaction is,
\[3{\rm{CaO}} + {\rm{2Al}} \to {\rm{3Ca + A}}{{\rm{l}}_2}{{\rm{O}}_{\rm{3}}}\]

The products are Ca and \[{\rm{A}}{{\rm{l}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}\] . Here, Ca is in solid state, so its enthalpy of formation is 0 and given that for \[{\rm{A}}{{\rm{l}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}\], \[\Delta {H_f} = - 1675\,{\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}\] .

\[{\Delta _r}{H^0}_f{\rm{(Products}}) = - 1675\,{\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ - 1}} + 3 \times 0 = - 1675\,{\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}\]
Therefore, the enthalpy of formation of products is -1675 \[{\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}\].

Now, we have to calculate the enthalpy of the formation of reactants. The reactants are Calcium oxide (CaO) and aluminium (Al). As aluminium is present in the state of solid, its enthalpy is zero. And given that, enthalpy of formation of calcium oxide is -635 \[{\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}\].

So, now, we will calculate the enthalpy of reactants.
\[{\Delta _r}{H^0}_f{\rm{(Reactants}}) = 3 \times - 635 + 2 \times 0 = - 1905\,{\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}\]

Now, we will calculate the standard enthalpy.
\[{\Delta _f}{H^0}_f = {\Delta _r}{H^0}_f{\rm{(Products}}) - {\Delta _r}{H^0}_f{\rm{(Reactants}})\]
\[{\Delta _r}{H^0} = - 1675\,{\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ - 1}} - \left( { - 1905\,{\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}} \right) = 230\,{\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}\]
Therefore, the standard enthalpy is \[230\,{\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}\].

Note: It is to be noted that, in the case of liquid and solid reactants, no appreciable volume change is observed. Therefore, a change in enthalpy equates to a change in internal energy. Standard enthalpy of formation also defines the formation of one mole of compound from its elements at standard conditions. The standard pressure is 1 atmosphere. In the SI unit, 1 bar is equal to \[{10^5}\] pascal and the standard temperature is 273 K.