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)
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
To determine the change in enthalpy for each reaction, you can use the standard enthalpies of formation of the reactants and products. The change in enthalpy (∆H) can be calculated using the following equation:
∆H = Σn(products)∆Hf(products) - Σn(reactants)∆Hf(reactants)
Where:
∆H = Change in enthalpy
Σn = Stoichiometric coefficient of the reactant or product
∆Hf = Standard enthalpy of formation
Let's calculate the change in enthalpy for each reaction:
1) NaOH(s) + HCl(g) → NaCl(s) + H2O(g)
∆H = (1)(-411.0) + (1)(-92.3) - (1)(-426.7) - (1)(-241.8)
∆H = -411.0 - 92.3 + 426.7 + 241.8
∆H = 165.2 kJ/mol
The change in enthalpy for this reaction is 165.2 kJ/mol.
2) 2CO(g) + O2(g) → 2CO2(g)
∆H = (2)(-110.5) + (1)(0) - (2)(-393.5)
∆H = -221.0 + 0 + 787.0
∆H = 566.0 kJ/mol
The change in enthalpy for this reaction is 566.0 kJ/mol.
3) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
∆H = (1)(-74.8) + (1)(0) - (1)(-393.5) - (2)(-285.8)
∆H = -74.8 + 0 + 393.5 + 571.6
∆H = 890.3 kJ/mol
The change in enthalpy for this reaction is 890.3 kJ/mol.
4) 2H2S(g) + 3O2(g) → 2H2O(l) + 2SO2(g)
∆H = (2)(-20.1) + (3)(0) - (2)(-285.8) - (2)(-296.1)
∆H = -40.2 + 0 + 571.6 + 592.2
∆H = 1123.6 kJ/mol
The change in enthalpy for this reaction is 1123.6 kJ/mol.
5) 2NO(g) + O2(g) → 2NO2(g)
∆H = (2)(90.4) + (1)(0) - (2)(33.9)
∆H = 180.8 + 0 - 67.8
∆H = 113.0 kJ/mol
The change in enthalpy for this reaction is 113.0 kJ/mol.
To determine the change in enthalpy for each reaction, you need to use the standard enthalpies of formation provided in the table. The change in enthalpy, ΔH, can be calculated using the equation:
ΔH = Σ(nΔHf(products)) - Σ(nΔHf(reactants))
where ΔHf is the standard enthalpy of formation for each species, n is the stoichiometric coefficient of each species in the balanced equation, and Σ represents the sum of all the terms.
Let's go through each reaction and calculate the change in enthalpy:
1) NaOH(s) + HCl(g) → NaCl(s) + H2O(g)
ΔH = [(1)(-411.0) + (1)(-92.3)] - [(1)(-569.0) + (1)(-285.8)]
= -503.3 + 854.8
= 351.5
The change in enthalpy for this reaction is 351.5 kJ.
2) 2CO(g) + O2(g) → 2CO2(g)
ΔH = [(2)(-110.5) + (1)(0)] - [(2)(-393.5)]
= -221.0 + 787.0
= 566.0
The change in enthalpy for this reaction is 566.0 kJ.
3) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
ΔH = [(1)(-74.8) + (2)(0)] - [(1)(-393.5) + (2)(-285.8)]
= -74.8 + 815.1
= 740.3
The change in enthalpy for this reaction is 740.3 kJ.
4) 2H2S(g) + 3O2(g) → 2H2O(l) + 2SO2(g)
ΔH = [(2)(-20.1) + (3)(0)] - [(2)(-285.8) + (2)(-296.1)]
= -40.2 + 1143.0
= 1102.8
The change in enthalpy for this reaction is 1102.8 kJ.
5) 2NO(g) + O2(g) → 2NO2(g)
ΔH = [(2)(90.4) + (1)(0)] - [(2)(0) + (2)(33.9)]
= 180.8 - 67.8
= 113.0
The change in enthalpy for this reaction is 113.0 kJ.
Therefore, the change in enthalpy for each of the given reactions is as follows:
1) ΔH = 351.5 kJ
2) ΔH = 566.0 kJ
3) ΔH = 740.3 kJ
4) ΔH = 1102.8 kJ
5) ΔH = 113.0 kJ