Carbocation
I guess pls tell me if it wrong
Are produced along with a large quantitu of heat
Answer:
The equilibrium constant for the reversible reaction = 0.0164
Explanation:
At equilibrium the rate of forward reaction is equal to the rate of backwards reaction.
The reaction is given as
A ⇌ B
Rate of forward reaction is first order in [A] and the rate of backward reaction is also first order in [B]
The rate of forward reaction = |r₁| = k₁ [A]
The rate of backward reaction = |r₂| = k₂ [B]
(Taking only the magnitudes)
where k₁ and k₂ are the forward and backward rate constants respectively.
k₁ = 0.010 s⁻¹
k₂ = 0.0610 s⁻¹
|r₁| = 0.010 [A]
|r₂| = 0.016 [B]
At equilibrium, the rate of forward and backward reactions are equal
|r₁| = |r₂|
k₁ [A] = k₂ [B] (eqn 1)
Note that equilibrium constant, K, is given as
K = [B]/[A]
So, from eqn 1
k₁ [A] = k₂ [B]
[B]/[A] = (k₁/k₂) = (0.01/0.0610) = 0.0163934426 = 0.0164
K = [B]/[A] = (k₁/k₂) = 0.0164
Hope this Helps!!!
Answer:
"0.053457 M" of sulfuric acid.
Explanation:
The given values are:
= 10 mL solution
= 12.20 mL
= 22.20 mL
then,
M 0.103 M of NaOH,
= experiment will not be affected
= 10.38 mL
Now,
⇒ mol of NAOH = MV
= 
= 
Whether Sulfuric acid, then
⇒ 
⇒ 
⇒ 
Before any dilution:
⇒ 
(Sulfuric acid)
