Question

0 points Given the values below, determine the absolute entropy of the compound ammonium chloride, which can be produced from a reaction of ammonia (NH) and hydrogen chloride (HCI). Compound AH" (kJmole) AG", CkJ/mole) SJK mole) NH, - 46.19 - 16.66 192.5 HC (8) - 92.30 -95.27 186.69 NH,CĐ) - 203.0

Answer 1

Answer – We are given,

Reaction between NH3 and HCl –

NH3(g) + HCl(g) ------> NH4Cl(s)

The values are given for standard enthalpy, standard free energy and absolute entropy and we asked for the absolute entropy for ammonium chloride.

To determine the absolute entropy of ammonium chloride, we need to calculate the standard entropy of reaction. The standard entropy of reaction is determine from the standard enthalpy and standard free energy as follows.

Standard enthalpy of reaction –

We know,

ΔHorxn = ∑ ΔHof of product – ∑ ΔHof of reactant

= [ (ΔHof NH4Cl(s)] - [ ΔHof NH3(g) + ΔHof HCl(g)]

= (-314.4 kJ/mol) – ( -46.19 kJ/mol +(-92.30 kJ/mol) )

= -175.91 kJ

Standard Free energy

ΔGorxn = ∑ ΔGof of product – ∑ ΔGof of reactant

= [ (ΔGof NH4Cl(s)] - [ ΔGof NH3(g) + ΔGof HCl(g)]

= (-203.0 kJ/mol) – ( -16.66 kJ/mol +(-95.27 kJ/mol) )

= -91.07 kJ

The Temperature is 298.15 K

We know the formula,

ΔGo = ΔHo-T *ΔSo

By plugging the values,

-91.07 kJ = -175.91 kJ -29815 K * ΔSo

ΔSo = (-91.07 kJ +175.91 kJ ) ÷ (-298.15 kJ)

         = - 0.284 kJ

1 kJ = 1000 J

0.284 = 284 J

ΔSo = -284 J

Now we need to determine the absolute entropy for NH4­Cl is

ΔSo = ∑ So of product – ∑ So of reactant

-284 J = = [ So NH4Cl(s)] - [So NH3(g) + So HCl(g)]

-284 J = [ So NH4Cl(s)] – (192.5 +186.69)

-284 = [ So NH4Cl(s)] -379.19

[ So NH4Cl(s)] = -284 +279.19

                       = 95.19 J/K.mol

Thus, absolute entropy for ammonium chloride is 95.19 J/K.mol