The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.
Answer:
Nitrobenzene is too deactivated (by the nitro group) to undergo a Friedel-Crafts alkylation.
Explanation:
The benzene ring in itself does not easily undergo electrophilic substitution reaction. Some groups activate or deactivate the benzene ring towards electrophilic substitution reactions.
-NO2 ia a highly deactivating substituent therefore, Friedel-Crafts alkylation of nitrobenzene does not take place under any conditions.
This reaction scheme is therefore flawed because Nitrobenzene is too deactivated (by the nitro group) to undergo a Friedel-Crafts alkylation.
Answer:
869 g Cl₂O
Explanation:
To find the theoretical yield of Cl₂O, you need to (1) convert moles SO₂ to moles Cl₂O (via mole-to-mole ratio from reaction coefficients) and then (2) convert moles Cl₂O to grams Cl₂O (via molar mass). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 3 sig figs to reflect the sig figs of the given amount (10.0 moles).
1 SO₂ (g) + 2 Cl₂ (g) ----> 1 SOCl₂ (g) + 1 Cl₂O (g)
Molar Mass (Cl₂O): 2(35.453 g/mol) + 15.998 g/mol
Molar Mass (Cl₂O): 86.904 g/mol
10.0 moles SO₂ 1 mole Cl₂O 86.904 g
------------------------ x ---------------------- x ------------------ = 869 g Cl₂O
1 mole SO₂ 1 mole
Answer:
Explanation:
Increasing Volume while maintaining constant pressure requires a proportional increase in Temperature so the gas pressure will be maintained as constant.
Consider...
V₁ = V₁ V₂ = 4V₁
T₁ = T₁ T₂ = ?
Charles Law => T ∝ V at constant P ... that is, increasing temperature generates a proportional increase in volume to maintain constant pressure.
Empirical Charles Law Relation is ...
V₁/T₁ = V₂/T₂ => T₂ = T₁(V₂/V₁) = T₁(4V₁/V₁) = 4T₁
Increasing Volume of a gas by 4 times requires a 4 times increase in absolute temperature in order to maintain constant pressure.
