<u>Answer:</u> The value of
of the reaction is 28.38 kJ/mol
<u>Explanation:</u>
For the given chemical reaction:
![SO_2(g)+Cl_2(g)\rightarrow SO_2Cl_2(g)](https://tex.z-dn.net/?f=SO_2%28g%29%2BCl_2%28g%29%5Crightarrow%20SO_2Cl_2%28g%29)
- The equation used to calculate enthalpy change is of a reaction is:
![\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28reactant%29%7D%5D)
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(SO_2Cl_2(g))})]-[(1\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(Cl_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_2Cl_2%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_2%28g%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28Cl_2%28g%29%29%7D%29%5D)
We are given:
![\Delta H^o_f_{(SO_2Cl_2(g))}=-364kJ/mol\\\Delta H^o_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H^o_f_{(Cl_2(g))}=0kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_f_%7B%28SO_2Cl_2%28g%29%29%7D%3D-364kJ%2Fmol%5C%5C%5CDelta%20H%5Eo_f_%7B%28SO_2%28g%29%29%7D%3D-296.8kJ%2Fmol%5C%5C%5CDelta%20H%5Eo_f_%7B%28Cl_2%28g%29%29%7D%3D0kJ%2Fmol)
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(1\times (-364))]-[(1\times (-296.8))+(1\times 0)]=-67.2kJ/mol=-67200J/mol](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-364%29%29%5D-%5B%281%5Ctimes%20%28-296.8%29%29%2B%281%5Ctimes%200%29%5D%3D-67.2kJ%2Fmol%3D-67200J%2Fmol)
- The equation used to calculate entropy change is of a reaction is:
![\Delta S^o_{rxn}=\sum [n\times \Delta S^o_f_{(product)}]-\sum [n\times \Delta S^o_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20S%5Eo_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20S%5Eo_f_%7B%28reactant%29%7D%5D)
The equation for the entropy change of the above reaction is:
![\Delta S^o_{rxn}=[(1\times \Delta S^o_{(SO_2Cl_2(g))})]-[(1\times \Delta S^o_{(SO_2(g))})+(1\times \Delta S^o_{(Cl_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20S%5Eo_%7B%28SO_2Cl_2%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20S%5Eo_%7B%28SO_2%28g%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20S%5Eo_%7B%28Cl_2%28g%29%29%7D%29%5D)
We are given:
![\Delta S^o_{(SO_2Cl_2(g))}=311.9J/Kmol\\\Delta S^o_{(SO_2(g))}=248.2J/Kmol\\\Delta S^o_{(Cl_2(g))}=223.0J/Kmol](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7B%28SO_2Cl_2%28g%29%29%7D%3D311.9J%2FKmol%5C%5C%5CDelta%20S%5Eo_%7B%28SO_2%28g%29%29%7D%3D248.2J%2FKmol%5C%5C%5CDelta%20S%5Eo_%7B%28Cl_2%28g%29%29%7D%3D223.0J%2FKmol)
Putting values in above equation, we get:
![\Delta S^o_{rxn}=[(1\times 311.9)]-[(1\times 248.2)+(1\times 223.0)]=-159.3J/Kmol](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20311.9%29%5D-%5B%281%5Ctimes%20248.2%29%2B%281%5Ctimes%20223.0%29%5D%3D-159.3J%2FKmol)
To calculate the standard Gibbs's free energy of the reaction, we use the equation:
![\Delta G^o_{rxn}=\Delta H^o_{rxn}-T\Delta S^o_{rxn}](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5CDelta%20H%5Eo_%7Brxn%7D-T%5CDelta%20S%5Eo_%7Brxn%7D)
where,
= standard enthalpy change of the reaction =-67200 J/mol
= standard entropy change of the reaction =-159.3 J/Kmol
Temperature of the reaction = 600 K
Putting values in above equation, we get:
![\Delta G^o_{rxn}=-67200-(600\times (-159.3))\\\\\Delta G^o_{rxn}=28380J/mol=28.38kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D-67200-%28600%5Ctimes%20%28-159.3%29%29%5C%5C%5C%5C%5CDelta%20G%5Eo_%7Brxn%7D%3D28380J%2Fmol%3D28.38kJ%2Fmol)
Hence, the value of
of the reaction is 28.38 kJ/mol
Moles of Li2CO3 = 1.53/73.891 = 0.0207 mole
Since HCl is in excess, amount of CO2 will depend on the limiting reagent which is Li2CO3.
∴Moles of CO2 = Moles of Li2CO3 = 0.0207.
Answer: There are about 0.28 molecules in 43.9 g of carbon tetrachloride. If you are rounding up, it would be 0.3
Explanation: