Use the standard enthalpies of formation in the table below to determine the change in enthalpy for the reactions which follow the table.

CH4(g)= -74.8
CO2(g) = -393.5
CO(g = - 110.5
HCl(g = - 92.3
H2O(g) = -241.8
H2O(l) = -285.8
H2S(g) = - 20.1
H2SO4(l) = - 811.3
MgSO4(s) = - 1278.2
MnO(s) = -384.9
MnO2(s)= - 519.7
NaCl(s) = - 411.0
NaF(s) = - 569.0
NaHCO3(s)= - 947.7
NaOH(s)= - 426.7
NH3(g)= - 46.2
NH4Cl(s)= -315.4
NO(g)= + 90.4
NO2(g)= +33.9
SO2(g)= - 296.1
SO3(g)= - 395.2
SnCl4(1)= -545.2
SnO(s)= - 286.2
SnO2(s)= - 580.7
ZnO(s)= - 348.0
ZnS(s)= - 202.9

1)NaOH(s) + HCl(g) → NaCl(s) + H2O(g)
2)2CO(g) + O2(g) → 2CO2(g)
3)CH4(g) +202(g) → CO2(g) + 2H2O(1)
4)2H2S(g) + 302(g) → 2H2O(l) +2SO2 (g)
5)2NO(g) + O2(g) → 2NO2 (g)

1 answer

To determine the change in enthalpy for the reactions, you can use the enthalpies of formation to calculate the overall change in enthalpy.

1) NaOH(s) + HCl(g) → NaCl(s) + H2O(g)
ΔH = ΔHf(NaCl) + ΔHf(H2O) - ΔHf(NaOH) - ΔHf(HCl)
ΔH = (-411.0) + (-241.8) - (-426.7) - (-92.3)
ΔH = -411.0 + 241.8 + 426.7 - 92.3
ΔH = 165.2 kJ

2) 2CO(g) + O2(g) → 2CO2(g)
ΔH = 2*ΔHf(CO2) - 2*ΔHf(CO)
ΔH = 2*(-393.5) - 2*(-110.5)
ΔH = -787.0 + 221.0
ΔH = -566.0 kJ

3) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
ΔH = ΔHf(CO2) + 2*ΔHf(H2O) - ΔHf(CH4) - 2*ΔHf(O2)
ΔH = (-393.5) + 2*(-285.8) - (-74.8) - 2*(0)
ΔH = -393.5 - 571.6 + 74.8
ΔH = -890.3 kJ

4) 2H2S(g) + 3O2(g) → 2H2O(l) + 2SO2(g)
ΔH = 2*ΔHf(H2O) + 2*ΔHf(SO2) - 2*ΔHf(H2S) - 3*ΔHf(O2)
ΔH = 2*(-285.8) + 2*(-296.1) - 2*(-20.1) - 3*(0)
ΔH = -571.6 - 592.2 + 40.2
ΔH = -1123.6 kJ

5) 2NO(g) + O2(g) → 2NO2(g)
ΔH = 2*ΔHf(NO2) - 2*ΔHf(NO)
ΔH = 2*(33.9) - 2*(90.4)
ΔH = 67.8 - 180.8
ΔH = -113.0 kJ