Complete Question
Calculate enthalpy changes for the following:
a
0.047 g of sulfur (rhombic) burns, forming [tex]SO_2_{(g)}[/tex] ( [tex]\Delta _f H^o[/tex] for [tex]SO_2_{(g)}[/tex] = – 296.84 kJ/mol)
Enthalpy change =______________ kJ
b
0.22 mol of decomposes to and [tex]Hg_{(l)}[/tex] and [tex]O_2_{(g)}[/tex] ( [tex]\Delta_f H^o[/tex]for [tex]HgO[/tex] = –90.83
kJ/mol) Enthalpy change = ______________kJ
c
2.44 g of [tex]NH_{3}_{(g)}[/tex] is formed from and excess [tex]N_{2}_{(g)}[/tex] and excess [tex]H_{2}_{(g)}[/tex]( [tex]\Delta _f H^o[/tex]
for [tex]NH_{3}[/tex] = –45.90 kJ/mol)
Enthalpy change =___________________ kJ
d
[tex]1.47 *10^{-2}[/tex] mol of carbon is oxidized to [tex]CO_{2}_{(g)}[/tex] ( [tex]\Delta_f H^o[/tex] for [tex]CO_2[/tex] = –393.509 kJ/mol)
Enthalpy change =__________________ kJ
Answer:
a
[tex]\Delta E_1= -0.435kJ[/tex]
b
[tex]\Delta E_2 = -19.98 kJ[/tex]
c
[tex]\Delta E_3= -6.6096kJ[/tex]
d
[tex]\Delta E_4 = -5.784 kJ[/tex]
Explanation:
For a
The number o moles ([tex]n_1[/tex]) = [tex]\frac{mass}{Molar \ mass \ of \ sulfur } = \frac{0.047}{32} =0.0015 \ moles[/tex]
The Change in enthapy is mathematically represented as
[tex]\Delta E_1 =n * \Delta _f H^o[/tex]
Substituting values
[tex]\Delta E_1 = 0.0015 * -296.84[/tex]
[tex]\Delta E_1= -0.435kJ[/tex]
For b
The Change in enthapy is mathematically represented as
[tex]\Delta E_2 =n_2 * \Delta _f H^o[/tex]
Substituting values
[tex]\Delta E_2 = 0.22 * -90.83[/tex]
[tex]\Delta E_2 = -19.98 kJ[/tex]
For c
The number o moles ([tex]n_3[/tex]) = [tex]\frac{mass}{Molar \ mass \ of \ NH_{3}_{(g)} } = \frac{2.44}{17} =0.144 \ moles[/tex]
The Change in enthapy is mathematically represented as
[tex]\Delta E_3 =n_3 * \Delta _f H^o[/tex]
Substituting values
[tex]\Delta E_3 = 0.144 * -45.90[/tex]
[tex]\Delta E_3= -6.6096kJ[/tex]
For d
The Change in enthapy is mathematically represented as
[tex]\Delta E_4 =n_4 * \Delta _f H^o[/tex]
Substituting values
[tex]\Delta E_4 = 1.47*10^{-2} * -393.509[/tex]
[tex]\Delta E_4 = -5.784 kJ[/tex]
