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

Step-by-step 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* \Delta H^o_f_((product))]-\sum [n* \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* \Delta H^o_f_((SiC(s))))+(2* \Delta H^o_f_((CO(g))))]-[(1* \Delta H^o_f_((SiO_2(s))))+(3* \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* \Delta H^o_f_((SiC(s))))+(2* (-110.5))]-[(1* (-910.9))+(3* (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.