Answer:
Explanation:
[tex]\text{From the information given:}[/tex]
[tex]\text{The chemical reaction is : } 4 KClO_{3(s)} \to 3 KClO_{4(s)} + KCl_{(s)}[/tex]
[tex]\text{To find} \ \Delta G^0_{rxn}\ \text{using the formula}: \\ \\ \Delta G^0_{rxn} = \sum n_p \times \Delta _f G^0 (Products) - \sum n_R \times \Delta _fG^0 ( Reactants) \\ \\ where; n_p = \text{no of moles of products } \ and; \\ \\ n_R = \text{no of moles of reactants }[/tex]
[tex]\implies G^0_{rxn} = 3 \times \Delta _fG^0 [KClO_4{(s)}] + \Delta_fG^0[KCl_{(s)}] - 4 \times \Delta _f G^0 [ KClO_3 (s) ][/tex]
[tex]\Delta _fG^0 \ values \ at \ 25^0 \ C (298 \ K) are\ given \ as:\\\\ \Delta _fG^0 [KClO_4(s)] = -303.09 \ kJ \\ \\ \Delta _fG^0 [KCl(s) ] = - 409.14 \ kJ \\ \\ \Delta_f G^0 [KClO_3_{(s)}] = -296.25 \ kJ \\ \\ replacing \ the \ above \ values \ into \ equation (1) ; then:\\ \\ \\\Delta G^0_{rxn} = 3 *(-303.09) + (-409.14) - 4*(-296.25) \ kJ \\ \\ = (-909.27 - 409.14 + 1185) \ kJ \\ \\ = -133.41 \ kJ \\ \\ \mathbf{\Delta G^0_{rxn} = -133.4 \ kJ }[/tex]
The standard free energy change of the reaction is -133 kJ.
From the reaction equation, we have; 4KClO3 ⇄ 3KClO4 (s) + KCl (s). The standard free energy of formation of each specie is given below;
ΔG°f KClO3 = -296.35 kJ
ΔG°f KClO4 = -303.09 kJ
ΔG°f KCl = -409.14 kJ
Hence;
ΔG°rxn = [3(-303.09)] + ( -409.14)] - [(4( -296.35))]
ΔG°rxn = (-909.27) + (-409.14) - (-1185.4)
ΔG°rxn = -133 kJ
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