Question

Consider the following equation: SiO2 (s) + 3C (graphite) --> SiC (s) + 2CO (g) ΔH rxn = 624.6 kJ / mol rxn. Using the following standard enthalpy of formation data, calculate standard enthalpy of formation for SiC (s). a. standard enthalpy of formation SiO2 (s) = -910.9 kJ/mol b. standard enthalpy of formation CO (g) = -110.5 kJ/mol

Answer 1

Answer: The enthalpy of the formation of $SiC(s)$ is coming out to be -65.3 kJ/mol

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:  

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

For the given chemical reaction:

$SiO_2(s)+3C\text{ (graphite)}(s)\rightarrow SiC(s)+2CO(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(SiC(s))})+(2\times \Delta H^o_f_{(CO(g))})]-[(1\times \Delta H^o_f_{(SiO_2(s))})+(3\times \Delta H^o_f_{(C(s))})]$

We are given:

$\Delta H^o_f_{(CO(g))}=-110.5kJ/mol\\\Delta H^o_f_{(SiO_2(s))}=-910.9kJ/mol\\\Delta H^o_f_{(C(s))}=0kJ/mol\\\Delta H^o_{rxn}=624.6kJ$

Putting values in above equation, we get:

$624.6=[(1\times \Delta H^o_f_{(SiC(s))})+(2\times (-110.5))]-[(1\times (-910.9))+(3\times (0))]\\\\\Delta H^o_f_{(SiC(s))}=-65.3kJ/mol$

Hence, the enthalpy of the formation of $SiC(s)$ is coming out to be -65.3 kJ/mol.