Answer:
D. The electrochemical reaction of the battery must be reversible.
Explanation:
The batteries are based on the production of an electric flux given by a<u> redox reaction</u>. This reaction is <u>spontaneous</u> and is<u> thermodynamically favored</u>.
Thus, when the reactants to the reaction are finished, the flow of current stops and ends. Therefore, when current is administered from another source, the reaction <u>changes its direction</u> and reagents that were previously consumed begin to occur. Therefore the condition for it to be <u>rechargeable</u> is that the reaction can go <u>forward or backward</u>, that is, it is <u>reversible</u>.
Answer:
false it's 24 hours long.
<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
The molar mass of methylammonium bromide is 111u.
<h3>What is molar mass?</h3>
The molar mass is defined as the mass per unit amount of substance of a given chemical entity.
Multiply the atomic weight (from the periodic table) of each element by the number of atoms of that element present in the compound.
Add it all together and put units of grams/mole after the number.
Atomic weight of H is 1u
Atomic weight of N is 14u
Atomic weight of C is 12u
Atomic weight of Br is 79u
Calculating molar mass of
=2(1 x3+ 14+12+ 1 x 3 +79) = 111u
Hence, the molar mass of methylammonium bromide is 111u.
Learn more about molar mass here:
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