NH4Cl(s) → NH3(g) + HCl(g) ∆H4 = ? Solid ammonium chloride sublimes (i.e., changes from a solid to a gas) at 340oC and decomposes to gaseous ammonia and hydrogen chloride. ∆H4 is obtained via mathematical manipulation of four equations and the enthalpy changes corresponding to the four related reactions: (5) NH3(aq) + HCl(aq) → NH4Cl(aq) ∆H5 (measured) (6) NH4Cl(s) → NH4Cl(aq) ∆H6 (measured) (7) NH3(g) → NH3(aq) ∆H7 = -34.64 kJ/mol (8) HCl(g) → HCl(aq) ∆H8 = -75.13 kJ/mol Select one: a. ΔH4 = ΔH6 - ΔH5 + ΔH7 + ΔH8 b. ΔH4 = - ΔH6 + ΔH5 + ΔH7 + ΔH8 c. ΔH4 = ΔH6 + (2xΔH5) + ΔH7 - ΔH8 d. ΔH4 = ΔH6 - ΔH5 - ΔH7 - ΔH8 e. None of the above answers are correct
Using the equation labels in the laboratory procedure, which of the following illustrates the correct calculation for ΔH4?
(1) C(s) + 1/2O2(g) → CO(g) ∆H1 = ?
(2) C(s) + O2(g) → CO2(g) ∆H2 = -395.7 kJ/mol
(3) CO(g) + 1/2O2(g) → CO2(g) ∆H3 = -283.2 kJ/mol
Summing reactions appropriately to determine ∆H1:
C(s) + O2(g) → CO2(g) ∆H2 = -395.7 kJ/mol (Reaction 2)
CO2(g) → CO(g) + 1/2O2(g) (-∆H3) = 283.2 kJ/mol (reverse of 3)
Sum: C(s) + 1/2O2 → CO(g) -395.7 + 283.2 = -112.5 kJ/mol
Determination of the Enthalpy of Sublimation and Decomposition of NH4Cl
In the laboratory experiment today, the heat of sublimation and decomposition of NH4Cl(s) will be determined. The process is represented by Equation 4.
(4) NH4Cl(s) → NH3(g) + HCl(g) ∆H4 = ?
Solid ammonium chloride sublimes (i.e., changes from a solid to a gas) at 340oC and decomposes to gaseous ammonia and hydrogen chloride. ∆H4 is obtained via mathematical manipulation of four
equations and the enthalpy changes corresponding to the four related reactions:
(5) NH3(aq) + HCl(aq) → NH4Cl(aq) ∆H5 (measured)
(6) NH4Cl(s) → NH4Cl(aq) ∆H6 (measured)
(7) NH3(g) → NH3(aq) ∆H7 = -34.64 kJ/mol
(8) HCl(g) → HCl(aq) ∆H8 = -75.13 kJ/mol
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