Q6E
Question
The reaction of quicklime,\({\rm{CaO}}\), with water produces slaked lime, \({\rm{Ca}}{({\rm{OH}})_2}\), which is widely used in the construction industry to make mortar and plaster. The reaction of quicklime and water is highly exothermic: \({\rm{CaO}}(s) + {{\rm{H}}_2}{\rm{O}}(l) \to {\rm{Ca}}{({\rm{OH}})_2}(s)\;\;\;\Delta H = - 350\;{\rm{kJ}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}\)
(a) What is the enthalpy of reaction per gram of quicklime that reacts?
(b) How much heat, in kilojoules, is associated with the production of 1 ton of slaked lime?
Step-by-Step Solution
Verified(a) Enthalpy of reaction,\({\rm{\Delta }}{H_{{\rm{mass\;}}}} = - 6.24kJ/g\).
(b) Heat in Kilojoules,\(q = - 6.24 \times {10^6}kJ\).
Enthalpy of reaction depicts the change in heat during the transformation of matter keeping all the reactants and products at their standard conditions.It is also termed as the heat of reaction.
(a)
If a certain reaction has a certain enthalpy per mole of a reactant, calculate the mass of one mole of the reactant (here\({\rm{CaO}}\)) to convert the unit.
\(\begin{aligned}{{}{\underline{\phantom{xx}}}}{m(CaO) = n(CaO) \times M(CaO)}\\\begin{aligned}{\underline{\phantom{xx}}}m(CaO) = 1mol \times 56.08g/mol\\m(CaO) = 56.08g\end{aligned}\\{}\\{{\rm{ \Delta }}{H_{mass}} = - 350\frac{{{\rm{kJ}}}}{{mol}} \times \frac{{1mol}}{{56.08g}} = - 6.24kJ/g}\end{aligned}\)
(b)
Released heat is the product of enthalpy (per mass) and the given mass. Thus, it can be represented as:
\(\begin{aligned}{{}{\underline{\phantom{xx}}}}{q = {\rm{\Delta }}{H_{{\rm{mass\;}}}} \times m}\\{{\rm{ }}q = - 6.24kJ/g \times 1{\rm{\;ton\;}} \times \frac{{1 \times {{10}^6}g}}{{1{\rm{\;ton\;}}}}}\\{{\rm{ }}q = - 6.24 \times {{10}^6}kJ}\end{aligned}\)