Use the information provided to determine _DrH° for the following reaction: $$\begin{array} { l c c } { \mathrm { D } _ { f } H ^ { \circ } \left( \mathrm { kJ } \mathrm { mol } ^ { - 1 } \right) } & \mathrm { CH } _ { 4 } ( g ) + 4 \mathrm { Cl } _ { 2 } ( g ) \rightarrow \mathrm { CCl } _ { 4 } ( g ) + 4 \mathrm { HCl } ( g ) & \mathrm { D } _ { \mathrm { r } } H ^ { \circ } = ? \ \mathrm { CH } _ { 4 } ( g ) & - 75 \ \mathrm { CCl } _ { 4 } ( g ) & - 96 \ \mathrm { HCl } ( g ) - 92 & \end{array}$$

Question

Quizplus · Accepted Answer

The enthalpy change for a reaction is equal to the difference in the enthalpies of formation of the products and reactants. In this case, the enthalpy of formation of the products is -96 kJ/mol for CCl4 and -92 kJ/mol for 4 HCl, for a total of -448 kJ/mol. The enthalpy of formation of the reactants is -75 kJ/mol for CH4 and 0 kJ/mol for 4 Cl2, for a total of -75 kJ/mol. Therefore, the enthalpy change for the reaction is -448 kJ/mol - (-75 kJ/mol) = -389 kJ/mol.

Use the Information Provided to Determine DrH° for the Following

Use the Information Provided to Determine _DrH° for the Following