Question

Potassium chlorate, a common oxidizing agent in fireworks and matchheads, undergoes a solid-state disproportionation reaction when heated:
4KClO3 (s) ⟶ Δ3KClO4 (s) + KCl (s).

Use ΔHf ° and S° values to calculate ΔG_sys ° (which is ΔGrxn °) in kJ at 25°C for this reaction.

Answer 1

Answer:

Explanation:

$\text{From the information given:}$

$\text{The chemical reaction is : } 4 KClO_{3(s)} \to 3 KClO_{4(s)} + KCl_{(s)}$

$\text{To find} \ \Delta G^0_{rxn}\ \text{using the formula}: \\ \\ \Delta G^0_{rxn} = \sum n_p \times \Delta _f G^0 (Products) - \sum n_R \times \Delta _fG^0 ( Reactants) \\ \\ where; n_p = \text{no of moles of products } \ and; \\ \\ n_R = \text{no of moles of reactants }$

$\implies G^0_{rxn} = 3 \times \Delta _fG^0 [KClO_4{(s)}] + \Delta_fG^0[KCl_{(s)}] - 4 \times \Delta _f G^0 [ KClO_3 (s) ]$

$\Delta _fG^0 \ values \ at \ 25^0 \ C (298 \ K) are\ given \ as:\\\\ \Delta _fG^0 [KClO_4(s)] = -303.09 \ kJ \\ \\ \Delta _fG^0 [KCl(s) ] = - 409.14 \ kJ \\ \\ \Delta_f G^0 [KClO_3_{(s)}] = -296.25 \ kJ \\ \\ replacing \ the \ above \ values \ into \ equation (1) ; then:\\ \\ \\\Delta G^0_{rxn} = 3 *(-303.09) + (-409.14) - 4*(-296.25) \ kJ \\ \\ = (-909.27 - 409.14 + 1185) \ kJ \\ \\ = -133.41 \ kJ \\ \\ \mathbf{\Delta G^0_{rxn} = -133.4 \ kJ }$