Question

Consider the following thermochemical equations. PCl5 (s)→PCl3 (g)+Cl2 (g)2P (s)+3Cl2 (g)→2PCl3 (g)ΔH∘rxn=87.9kJmol ΔH∘rxn=−574kJmol Using this data, determine the heat of formation for PCl5.

Answer 1

Answer:

• -662 kJ/mol

Explanation:

The standard heat of formation is the enthalpy change for the formation of a compound from the elements at their standard states.

PCl₅ is formed by P and Cl, whose standard states are solid and diatomic gas.

You just must manipulate the two given equations to obtain an overall equation from P(s) and Cl₂(g).

Based of Hess' law, the enthalpy change of the overall equation is the algebraic sum of the individual equations.

If you multiply the first equation by 2 and reverse it you get:

     $2Cl_2(g)+2PCl_3(g)\rightarrow 2PCl_5(s);\Delta H^0rxn=-2mol\times 87.9kJ/mol$

Write the second equation:

    $2P(s)+3Cl_2)(g)\rightarrow 2PCl_3(g);\Delta H^0rxn=2mol\times (-574)kJ/mol$

Add them:

      $2P(s)+5Cl_2(g)\rightarrow 2PCl_5(s)$

                                      $\Delta H^0rxn=-2mol\times 87.9kJ/mol-2mol\times 574kJ/mol$

After computing:

        $\Delta H^0rxn=-1,323.8kJ$

Notice that the last calculation results in the standard enthalpy of formation for two moles of PCl₅(s). Thus, to obtain the enthalpy of formation per mol you must divide by 2:

       $\Delta H^0rxn=-661.9kJ/mol$

Round to 3 signficant figures: -662kJ/mol