<u>Answer:</u> The 
for the reaction is 54.6 kJ/mol
<u>Explanation:</u>
For the given balanced chemical equation:
We are given:
- To calculate
for the reaction, we use the equation:
![\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G_f%28reactant%29%5D)
For the given equation:
![\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28COCl_2%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28CO_2%29%7D%29%2B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28CCl_4%29%7D%29%5D)
Putting values in above equation, we get:
![\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%28-204.9%29%29-%28%281%5Ctimes%20%28-394.4%29%29%2B%281%5Ctimes%20%28-62.3%29%29%29%5D%5C%5C%5CDelta%20G%5Eo_%7Brxn%7D%3D46.9kJ%3D46900J)
Conversion factor used = 1 kJ = 1000 J
- The expression of
for the given reaction:
We are given:
Putting values in above equation, we get:
- To calculate the Gibbs free energy of the reaction, we use the equation:
where,
= Gibbs' free energy of the reaction = ?
= Standard gibbs' free energy change of the reaction = 46900 J
R = Gas constant = 
T = Temperature = ![25^oC=[25+273]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B25%2B273%5DK%3D298K)
= equilibrium constant in terms of partial pressure = 22.92
Putting values in above equation, we get:
Hence, the for the reaction is 54.6 kJ/mol
Answer:
ΔH = - 44.0kJ
Explanation:
H2O(l)→H2O(g), ΔH =44.0kJ
In the reaction above, liquid water changes to gaseous water. This occurs through a process known as boiling. This process requires heat, hence the ΔH is positive.
If he reaction is reversed, we have;
H2O(g)→H2O(l)
In this reaction, gaseous water changes to liquid water. This process is known as condensation. The water vapor loses heat in this reaction. Hence ΔH would be negative but still have the same value.
Answer:
a. withdraws electrons inductively
b. donates electrons by hyperconjugation
c. donates electrons by resonance
d. withdraws electrons inductively
Explanation:
a. The bromide ion is a highly electronegative ion (in the halide series). Electronegative substituents on acids increase the acidity by inductive electron withdrawal method. The higher the electronegativity of a substance, the greater the acidity. The halogens have this order of electronegativity:
F > Cl > Br>I
b. The carboxyl groups have a stabilization of the sigma and pi bonds. This is achieved through a special delocalization of electrons. Because of the delocalization, hyperconjugation is the result effect.
c. The NHCH₃ group has a highly electonegative nitrogen atom that pulls the electron cloud towards itself. In this case, it withdraws electrons inductively. As a result, it donates electrons by resonance.
d. The OCH₃ group has a highly electonegative oxygen atom. This oxygen atom withdraws electron cloud towards itself. As a result, it withdraws electrons inductively.
