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kondaur [170]
3 years ago
7

In a different experiment, the student uses a calorimeter which is perfectly insulated. She fills the calorimeter with 100.0 g o

f 25.00°C dilute HCl solution and adds 0.594 g Mg metal (24.3 g/mol). The final temperature of the apparatus comes to 41.83°C. Remember, the solution is made-up of the solute and the solvent. The specific heat for magnesium is negligible compared to the aqueous solution. What is the Molar Heat of Enthalpy for this reaction?
Chemistry
1 answer:
alexdok [17]3 years ago
8 0

Answer:

Explanation:

Mg + 2HCl = Mg Cl₂ + H₂

.594 g = .594 / 24.3

= .02444 mole

Heat evolved = msΔ T , m is mass of water ( solvant ) , s is specific heat of water , Δ T is rise in temperature

= 100 x 4.2 x ( 41.83 - 25 )

= 7068.6 J

.02444 mole  of Mg evolves 7068.6 J of heat

1 mole of Mg evolves 7068.6 /.02444 J

= 289222.6 J

= 289 kJ .

Molar heat enthalpy = 289 kJ .

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\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]

The equation for the Gibbs free energy change of the above reaction is:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]

We are given:

\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ

Putting values in above equation, we get:

285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol

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