Question

The most important contributor to the strength of hardened Portland cement is tricalcium silicate, ${\left(\mathbf{CaO}\right)}_3\mathbf{.}{\mathbf{SiO}}_2$ , for which the measured standard enthalpy of formation is $\mathbf{-}\mathbf{2929}\mathbf{.}\mathbf{2}\mathbf{kJ}\mathbf{.}{\mathbf{mol}}^{\mathbf{-}\mathbf{1}}$ . Calculate the standard enthalpy change for the production of 1.00 mol tricalcium silicate from quartz and lime.

Short answer

The standard enthalpy of formation of 1 mol. of tricalcium silicate is -2233.5 kJ/mol.

Step-by-step solution

Standard enthalpy change of a reaction

The standard enthalpy change for a chemical reaction can be calculated by using the below shown formula:

${\mathbf{\Delta H}}_{\mathrm{rxn}}\mathbf{=}\sum {\mathbf{\Delta H}}_{\mathrm{products}}\mathbf{-}\sum {\mathbf{\Delta H}}_{\mathrm{reactants}}$

The chemical equation involved in the given reaction

The product is tricalcium silicate.

The reactants are quartz ${\mathrm{SiO}}_2$ and lime ${\mathrm{CaCO}}_3$.

The balanced chemical equation is as shown below:

$3{\mathrm{CaCO}}_3\left(\mathrm{s}\right)+{\mathrm{SiO}}_2\left(\mathrm{s}\right)\to {\left(\mathrm{CaO}\right)}_3.{\mathrm{SiO}}_2\left(\mathrm{g}\right)+3{\mathrm{CO}}_2\left(\mathrm{g}\right)$

Calculation

Substitute the standard enthalpy of formation of each substrate molecule in the above formula to get standard enthalpy of formation of tricalcium silicate:

$\begin{array}{c}{\mathrm{\Delta H}}_{\mathrm{rxn}}=[{\mathrm{\Delta H}}_{{\left(\mathrm{CaO}\right)}_3.{\mathrm{SiO}}_2}+{\mathrm{\Delta H}}_{{\mathrm{CO}}_2}]-[{\mathrm{\Delta H}}_{{\mathrm{SiO}}_2}+{\mathrm{\Delta H}}_{{\mathrm{CaCO}}_3}]\\ =(-2929.2-393.5)\mathrm{kJ}/\mathrm{mol}-(-910.9-178.3)\mathrm{kJ}/\mathrm{mol}\\ =-2233.5\mathrm{kJ}/\mathrm{mol}\end{array}$

The negative sign indicates that the given reaction is exothermic.