which resonance form would be the most stable contributor to the intermediate arenium ion produced by electrophilic bromination
of methoxybenzene
1 answer:
The question is incomplete as the options are missing, however, the correct complete question is attached.
Answer:
The correct answer is option A. ( check image)
Explanation:
The most stable contributor to the intermediate arenium ion produced by electrophilic bromination of methoxybenzene in given options is option a due to the fact that this resonating form follows the octet rule is satisfied for all atoms and additional π bond is present in between C-O that makes it more stable, while in other options there are positive charge which means they do not follows octet rule completely.
Thus, the correct answer is option A. ( check image)
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Answer:
The correct option is <em>b) SO₃(g) + NO(g) ⇌ SO₂(g) + NO₂(g)</em>
Explanation:
The relation between Kc and Kp is given by the following equation:
where R is the gas constant (0,082 L.atm/K.mol), T is the temperature (in K) and dn is the change in moles.
The change in moles (dn) is calculated as:
dn = moles of products - moles reactants
If dn=0, RT= 1 ⇒ Kc=Kp
We calculate dn for each reaction from the estequiometrial coefficients of products and reactants as follows:
a) 4 NH₃(g) + 3 O₂(g) ⇌ 2 N₂(g) + 6 H₂O(g)
dn= (2+6) - (4+3) = 1 ⇒ Kc ≠ Kp
b) SO₃(g) + NO(g) ⇌ SO₂(g) + NO₂(g)
dn = (1+1) - (1+1)= 0 ⇒ Kc = Kp
c) 2 N₂(g) + O₂(g) ⇌ 2 N₂O(g)
dn= 2 - (2+1) = -1 ⇒ Kc ≠ Kp
d) 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g)
dn = 2 - (2+1) = -1 ⇒ Kc ≠ Kp
The reaction in which Kc=Kp is b), because reactants and products have the same number of moles.