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PRACTICE PROBLEMS
Sample Problem
Calculate the enthalpy change for the reaction in which hydrogen gas,
H2 (g), is combined with fluorine gas, F2(g), to produce 2 moles of
hydrogen fluoride gas, HF(g). This reaction is represented by the
balanced chemical equation
H2(g)+F2(g) → 2HF(g)
Given:
for H2(g): nH–H = 1 mol; DH–H = 432 kJ/mol;
for F2(g): nF–F = 1 mol; DF–F = 154 kJ/mol;
Required:ΔH
Analysis: ∆H = Σn x D bonds broken –Σn x D bonds formed
ΔH = (nH-HDH-H + nF-F DF-F) – (nH-FDH-F
5.3
Solution:
1 mol each of H–H and F–F bonds are broken
The bonds formed are 2 mol of H–F bonds
∆H= (nH-HDH-H + nF-FDF-F) – nH-FDH-F
(1 mol x 432KJ) + (1 mol x 154 KJ) - (2 mol x 565 KJ
mol
mol
mol
∆H = -544 KJ
The enthalpy change for the reaction of 1 mol hydrogen gas
and 1 mol fluorine gas to ptoduce 2 mol. Hydrogen fluoride
is -544 KJ
5.3
5.5: Standard enthalpy of formation
Standard enthalpy of formation (DH0f) is the heat change
that results when one mole of a compound is formed from
its elements at a pressure of 1 atm.
The standard enthalpy of formation of any element in its most
stable form is zero.
DH0 (O2) = 0
f
DH0 (O3) = 142 kJ/mol
f
DH0f (C, graphite) = 0
DHf0 (C, diamond) = 1.90 kJ/mol
5.5
0 ) is the enthalpy of
The standard enthalpy of reaction (DHrxn
a reaction carried out at 1 atm.
aA + bB
cC + dD
DH0rxn = [ cDH0f (C) + dDH0f (D) ] - [ aDH0f (A) + bDH0f (B) ]
DH0rxn = S nDH0f (products) - S mDHf0 (reactants)
Hess’s Law: When reactants are converted to products, the
change in enthalpy is the same whether the reaction takes
place in one step or in a series of steps.
(Enthalpy is a state function. It doesn’t matter how you get
there, only where you start and end.)
5.5
Calculate the standard enthalpy of formation of CS2 (l)
given that:
C(graphite) + O2 (g)
CO2 (g) DH0rxn = -393.5 kJ
S(rhombic) + O2 (g)
CS2(l) + 3O2 (g)
SO2 (g)
DH0rxn = -296.1 kJ
CO2 (g) + 2SO2 (g)
0 = -1072 kJ
DHrxn
1. Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic)
CS2 (l)
2. Add the given rxns so that the result is the desired rxn.
C(graphite) + O2 (g)
2S(rhombic) + 2O2 (g)
+ CO2(g) + 2SO2 (g)
CO2 (g) DH0rxn = -393.5 kJ
2SO2 (g) DH0rxn = -296.1x2 kJ
CS2 (l) + 3O2 (g)
0 = +1072 kJ
DHrxn
C(graphite) + 2S(rhombic)
CS2 (l)
0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ
DH
rxn
5.5
Benzene (C6H6) burns in air to produce carbon dioxide and
liquid water. How much heat is released per mole of
benzene combusted? The standard enthalpy of formation
of benzene is 49.04 kJ/mol.
2C6H6 (l) + 15O2 (g)
12CO2 (g) + 6H2O (l)
DH0rxn = S nDH0f (products) - S mDHf0 (reactants)
DH0rxn = [ 12DH0f (CO2) + 6DH0f (H2O)] - [ 2DH0f (C6H6)]
DH0rxn = [ 12x–393.5 + 6x–285.8 ] – [ 2x49.04 ] = -6534.9 kJ
- 6534.9 kJ
= - 3267.44 kJ/mol C6H6
2 mol
5.5
Benzene (C6H6) burns in air to produce carbon dioxide and
liquid water. How much heat is released per mole of
benzene combusted? The standard enthalpy of formation
of benzene is 49.04 kJ/mol.
2C6H6 (l) + 15O2 (g)
12CO2 (g) + 6H2O (l)
DH0rxn = S nDH0f (products) - S mDHf0 (reactants)
DH0rxn = [ 12DH0f (CO2) + 6DH0f (H2O)] - [ 2DH0f (C6H6)]
DH0rxn = [ 12x–393.5 + 6x–285.8 ] – [ 2x49.04 ] = -6534.9 kJ
- 6534.9 kJ
= - 3267.44 kJ/mol C6H6
2 mol
5.5
The enthalpy of solution (DHsoln) is the heat generated or
absorbed when a certain amount of solute dissolves in a
certain amount of solvent.
DHsoln = Hsoln - Hcomponents
Which substance(s) could be
used for melting ice?
NaCl, KCl, NH4Cl, NH4NO3
Which substance(s) could be
used for a cold pack?
LiCl & CaCl2
6.6