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Calculate δg o for each reaction using δg of values:(a) h2(g) + i2(s) → 2hi(g) kj (b) mno2(s) + 2co(g) → mn(s) + 2co2(g) kj (c) nh4cl(s) → nh3(g) + hcl(g)

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Part (a) :
H₂(g) + I₂(s) → 2 HI(g)
From given table:
G HI = + 1.3 kJ/mol
G H₂ = 0
G I₂ = 0
ΔG = G(products) - G(reactants) = 2 (1.3) = 2.6 kJ/mol

Part (b):
MnO₂(s) + 2 CO(g) → Mn(s) + 2 CO₂(g)
G MnO₂ = - 465.2
G CO = -137.16
G CO₂ = - 394.39
G Mn = 0
ΔG = G(products) - G(reactants) = (1(0) + 2*-394.39) - (-465.2 + 2*-137.16) = - 49.3 kJ/mol

Part (c):
NH₄Cl(s) → NH₃(g) + HCl(g)
ΔG = ΔH - T ΔS
ΔG = (H(products) - H(reactants)) - 298 * (S(products) - S(reactants))
      = (-92.31 - 45.94) - (-314.4) - (298 k) * (192.3 + 186.8 - 94.6) J/K
      = 176.15 kJ - 84.78 kJ = 91.38 kJ 
 





 

(a) The change in Gibbs free energy for the reaction has been 2.6 kJ/mol.

(b) The change in Gibbs free energy for the reaction has been -49.3 kJ/mol.

(c) The change in Gibbs free energy for the reaction has been 91.38 kJ/mol.

[tex]\Delta G[/tex] has been defined as change in free energy of the system in a chemical reaction. The change in free energy has been expressed as:

[tex]\Delta G=\Delta G_{product}-\Delta G_{rectant}[/tex]

Computation for [tex]\Delta G[/tex] of reactions

The [tex]\Delta G[/tex] for the compounds has been given in the image attached below.

(a) [tex]\rm H_2\;+\;I_2\;\rightarrow\;2\;HI[/tex]

The value of [tex]\Delta G[/tex] has been given as:

[tex]\Delta G=\rm 2(HI)-(H_2\;+\;I_2)\\ \Delta \textit G=2(1.3)-(0\;+\;0)\;kJ/mol\\ \Delta \textit G=2.6\;kJ/mol[/tex]

The change in Gibbs free energy for the reaction has been 2.6 kJ/mol.

(b) [tex]\rm MNO_2\;+\;2\;CO\;\rightarrow\;Mn\;+\;2\;CO_2[/tex]

The value of [tex]\Delta G[/tex] has been given as:

[tex]\Delta G=\rm (Mn\;+\;2(CO_2))-(MnO_2\;+\;2(CO))\\\Delta \textit G=(0\;+\;2(394.39))-(-465.2\;+\;2(-137.16))\;kJ/mol\\\Delta \textit G=-49.3\;kJ/mol[/tex]

The change in Gibbs free energy for the reaction has been -49.3 kJ/mol.

(c) [tex]\rm NH_4\;\rightarrow\;NH_3\;+\;HCl[/tex]

The value of [tex]\Delta G[/tex] has been given as:

[tex]\Delta G=\Delta H-T\Delta S\\ \Delta G=\Delta H_{product}-\Delta H_{reactant}-T\;\Delta S_{product}-\Delta S_{reactant}\\ \Delta G= (-92.31 - 45.94) - (-314.4) - (298 \text K) \times (192.3 + 186.8 - 94.6) \text{J/K}\\ \Delta G=91.38\;\rm kJ[/tex]

The change in Gibbs free energy for the reaction has been 91.38 kJ/mol.

Learn more about change in Gibbs free energy, here:

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