Wastes
Fuel
Smoke
Gas
Toxic chemicals
Answer:
Explanation has been given below.
Explanation:
- Chloroform has three polar C-Cl bonds. Methylene chloride has two polar C-Cl bonds. So it is expected that chloroform should be more polar and posses higher dipole moment than methylene chloride.
- Two factors are liable for the opposite trend observed in dipole moments of methylene chloride and chloroform.
- First one is the number of hyperconjugative hydrogen atoms present in a molecule. Hyperconjugation occurs with vacant d-orbital of Cl atom. Hyperconjugation amplifies charge separation in a molecule resulting higher dipole moment.
- Methylene chloride has two hyperconjugative hydrogen atoms and chloroform has one hyperconjugative hydrogen atom.Therefore methylene chloride should have higher charge separation as compared to chloroform.
- Second one is induction of opposite polarity in a C-Cl bond by another C-Cl bond in a molecule. Higher the opposite induction of polarity, lower the charge separation in a molecule and hence lower the dipole moment of a molecule.
- Chloroform has three C-Cl bonds and methylene chloride has two C-Cl bonds. Therefore opposite induction is higher for chloroform resulting it's lower dipole moment.
Answer:
Sulfuric acid contains 2 hydrogen atoms, 1 sulfur atom, and 4 oxygen atoms.
Explanation:
Answer : The value of
for the reaction is -959.1 kJ
Explanation :
The given balanced chemical reaction is,

First we have to calculate the enthalpy of reaction
.

![\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2O%29%7D%2Bn_%7BSO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28SO_2%29%7D%5D-%5Bn_%7BH_2S%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2S%29%7D%2Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%7D%5D)
where,
= enthalpy of reaction = ?
n = number of moles
= standard enthalpy of formation
Now put all the given values in this expression, we get:
![\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5B2mole%5Ctimes%20%28-242kJ%2Fmol%29%2B2mole%5Ctimes%20%28-296.8kJ%2Fmol%29%7D%5D-%5B2mole%5Ctimes%20%28-21kJ%2Fmol%29%2B3mole%5Ctimes%20%280kJ%2Fmol%29%5D)

conversion used : (1 kJ = 1000 J)
Now we have to calculate the entropy of reaction
.

![\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28H_2O%29%7D%2Bn_%7BSO_2%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28SO_2%29%7D%5D-%5Bn_%7BH_2S%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28H_2S%29%7D%2Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28O_2%29%7D%5D)
where,
= entropy of reaction = ?
n = number of moles
= standard entropy of formation
Now put all the given values in this expression, we get:
![\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5B2mole%5Ctimes%20%28189J%2FK.mol%29%2B2mole%5Ctimes%20%28248J%2FK.mol%29%7D%5D-%5B2mole%5Ctimes%20%28206J%2FK.mol%29%2B3mole%5Ctimes%20%28205J%2FK.mol%29%5D)

Now we have to calculate the Gibbs free energy of reaction
.
As we know that,

At room temperature, the temperature is 500 K.


Therefore, the value of
for the reaction is -959.1 kJ