## Q. 6.PS.12

Thermochemical Expressions for Standard Molar Enthalpies of Formation

Rewrite thermochemical Expressions 6.9 and 6.10 so that they represent standard molar enthalpies of formation of their products. (The standard molar enthalpy of formation value for $MgSO_4(s)$ is – 1284.9 kJ/mol at 25 °C.)

## Verified Solution

$Mg( s ) + \frac{1}{2} S_8(s) + 2 O_2(g) → MgSO_4(s) ΔH^\circ_f \left\{MgSO_4( s )\right\} = – 1284.9 kJ/mol$

and

$\frac{1}{4} P_4(s) + \frac{3}{2}Cl_2(g) → PCl_3(\ell ) ΔH^\circ_f \left\{PCl_3(\ell )\right\} = – 319.7 kJ/mol$

Strategy and Explanation Expression 6.9 has 1 mol $MgSO_4(s)$ on the right side, but the reactants are not elements in their standard states. Write a new expression so that the left side contains the elements $Mg(s), S_8(s)$, and $O_2(g)$. For this expression $ΔH°$ is the standard molar enthalpy of formation, – 1284.9 kJ/mol. The new thermochemical expression is given in the Answer section above.
Expression 6.10 has elements in their standard states on the left side, but more than 1 mol of product is formed. Rewrite the expression so the right side involves only 1 mol $PCl_3(\ell )$, and reduce the coefficients of the elements on the left side in proportion—that is, divide all coefficients by 4. Then $ΔH°$ must also be divided by 4 to obtain the second thermochemical expression in the Answer section.

Reasonable Answer Check Check each expression carefully to make certain the substance whose standard enthalpy of formation you want is on the right side and has a coefficient of 1. For $PCl_3(\ell ), ΔH^\circ_f$ should be one fourth of about 1300 kJ, and it is.