**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.)

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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.

Question: 6.PS.14

ΔH_f^\circ =- 208.4 kJ
Strategy an...

Question: 6.PS.13

ΔH° = - 3267.5 kJ
Strategy and Exp...

Question: 6.PS.11

ΔH° = 137.0 kJ
Strategy and Explan...

Question: 6.PS.10

ΔH = - 59 kJ
Strategy and Explanat...

Question: 6.PS.9

- 2810 kJ
Strategy and Explanation When the glucos...

Question: 6.PS.8

- 335 kJ
Strategy and Explanation Begin by calcul...

Question: 6.PS.7

(a) \frac{1}{2}C_2H_6(g) + \frac{7}{4}O_2(...

Question: 6.PS.6

1.6 × 10^{11} kJ
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Question: 6.PS.5

ΔH = 5865 J; ΔE = 5427 J
Strate...

Question: 6.PS.4

T_{final} = 42.7 °C
Strategy and E...