H° rxn = nH f °(products) - mH f °(reactants) Compound/Ion H f ° (kJ/mol) Ca 2+
ID: 878779 • Letter: H
Question
H°rxn = nHf°(products) - mHf°(reactants)
Compound/Ion Hf° (kJ/mol)
Ca2+(aq)
-542.8
OH-(aq)
-230.0
Ca(OH)2 (s)
-985.2
Fe(OH)3 (s)
-824.25
H+(aq)
0.0
Fe3+(aq)
-48.5
H2O(l)
-285.8
Mg(OH)2(s)
-924.66
HF(aq)
-332.6
Mg2+(aq)
-466.9
F-(aq)
-332.6
Using the enthalpies of formation given above, compute the enthalpy change for each of the aqueous net ionic reactions given below.
a. Mg(OH)2(s) + 2HF(aq) Mg2+(aq) + 2F-(aq) + 2H2O(l)
b. 3Mg(s) + 2Fe3+(aq) 2Fe(s) + 3Mg2+(aq)
Ca2+(aq)
-542.8
OH-(aq)
-230.0
Ca(OH)2 (s)
-985.2
Fe(OH)3 (s)
-824.25
H+(aq)
0.0
Fe3+(aq)
-48.5
H2O(l)
-285.8
Mg(OH)2(s)
-924.66
HF(aq)
-332.6
Mg2+(aq)
-466.9
F-(aq)
-332.6
Explanation / Answer
Solution:
We can use following equation to calculate Enthalpy change for reaction (H°rxn)
H°rxn = nHf°(products) - mHf°(reactants)
Let’s consider first reaction (a)
a. Mg(OH)2(s) + 2HF(aq) Mg2+(aq) + 2F-(aq) + 2H2O(l)
H°rxn = [( Mg2+(aq) ) + (2F-(aq)) + (2H2O(l))] – [(Mg(OH)2) +(2HF(aq))
Let’s plug all above enthalpy value of individual ion into formula.
H°rxn = [(-466.9) + (2*-332.6) + (2*-285.8) ] – [(-924.66) + (2*-332.6)]
H°rxn = [(-466.9) + (-665.2) + (-571.6) ] – [(-924.66) + (-665.2)]
H°rxn = -113.84 KJ
Answer – Enthalpy change for reaction "a" is( H°rxn) = -113.84 KJ
Now let’s consider second reaction (b)
b. 3Mg(s) + 2Fe3+(aq) 2Fe(s) + 3Mg2+(aq)
H°rxn = nHf°(products) - mHf°(reactants)
H°rxn = [(2Fe(s)) + (3Mg2+(aq))] – [(3Mg(s)) +(2Fe3+(aq))
Let’s plug all above enthalpy value of individual ion into formula.
H°rxn = [(2 * 0) + (3 * -466.9)] – [(3 * 0) + (2 * -48.5)
(Here enthalpy of pure state element (Fe(s) & Mg(s)) = 0)
H°rxn = [0 +(-1400.7)] – [ 0 + (-97)]
H°rxn = -1303.7 KJ
Answer – Enthalpy change for reaction “b” is( H°rxn) = -1303.7 KJ
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