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Dhamnekar Winod
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General chemistry question

Post by Dhamnekar Winod »

How to compute \( \Delta H^\circ, \Delta S^\circ, \Delta G^\circ\) at \(25^\circ C\) for the following reactions?


1)Combustion of Methane

2)Formation of gaseous HCl from its elements.

3)Formation of \(H_3SiCl\) from its elements [ We will need to use bond energies and estimate \( \Delta S^\circ]\)

My Answers: I know the reaction equations for 1)\(CH_4 (g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)\)

2)\(Cl_2(g)+ H_2(g)\rightarrow 2HCl(g) \)
3) \( Si(s) + 3HCl(aq) \rightarrow H_3SiCl(g) + Cl_2(g)\)

\( \Delta H^\circ_{rxn} \)(Enthalpy of reaction) for first reaction is -393.5 kJ/mol -285.8 kJ/mol-(-74.6 kJ/mol)=-604.7 kJ/mol Reaction is exothermic.

\( \Delta H^\circ_{rxn} \) (Enthalpy of reaction) for second reaction is -92.3 kJ/mol Reaction is exothermic.

\( \Delta S^\circ_{rxn}\) Spontaneity for the first reaction is 213.8 J/mol K+ 140 J/mol K - 186.3 J/mol K -410.4 J/mol K = -242.9 J/mol K First reaction is spontaneous at low temperature.

\( \Delta S^\circ_{rxn}\) Spontaneity for the second reaction is 373.8 J/mol K - 223.1 J/mol K -130.7 J/mol K=20 J/mol K Second reaction is spontaneous at all temperature.

Gibbs free energy for the first reaction's \( ( \Delta G^\circ_r)\) is -604.7 kJ/mol - 298 k * (-242.9 J/mol K) =-532.32 kJ/mol

Gibbs free energy for the second reaction (\(\Delta G^\circ_r)\) is -92.3 kJ/mol- 298 K*(20.0 J/mol K)= -86.34 kJ/mol

Enthalpy of third reaction \((\Delta H^\circ_f)\) = -141.84 kJ/mol- (-501.6 kJ/mol)=359.76 kJ/mol This reaction is endothermic.

\(\Delta S^\circ_r\) for the third reaction is 250.76 J/mol K -188.3 J/mol K = 62.46 J/mol K The third reaction is non-spontaneous at low temperature.


\( \Delta G^\circ_r\) for the third reaction is 359.76 kJ/mol - 298 K*(62.46 J/mol K)=341.15 kJ/mol


Are my computations of \( \Delta H^\circ_f, \Delta S^\circ_r, \Delta G^\circ_r\) for all these three reactions correct?
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