Answer:
Use the Bromotriflouride catalyst, BF₃
Explanation:
The BF₃ is most likely to yield less desired side products. The effect lies in the reaction mechanism.
BF₃ is a Lewis acid. Its role is to promote the ionization of the HF. This is achieved through the electrophilic mechanism. The reaction mechanism is as follows:
2 - methylpropene + H-F-BF₃ → H-F + H₃C + benzene
butylbenzene + F-BF₃ → tert-butylbenzene + H-F + BF₃ (regenerated catalyst)
<span>Bases and Acids are chemically opposite from each other,and there are multiple ways to distinguish how they react when dissolved in water.
One accepted definition is that an acid is any chemical substance that, when it is dissolved in water, creates a solution with hydrogen ion activity greater than pure/neutral water. That is, it donates a proton to the solution. Any substance with a pH less than 7.0 is an acid, and includes substances such as vinegar and lemon juice.
By comparison, a base is any chemical substance that, when it is dissolved in water, creates a solution in which has hydrogen ion activity less than pure/neutral water. That is, it accepts protons. Any substance with a pH greater than 7.0 is a base, and includes substances such as ammonia and baking soda.</span>
Answer:
Q was < K. Partial pressure of hydrogen decreased, iodine increased
Explanation:
<em>After iodine was added the Q was [Select] K so the reaction shifted toward the Products [Select] ,The partial pressure of hydrogen [Select], Iodine [Select] |,and hydrogen iodide Decreased</em>
Based on the equilibrium:
H2(g) + I2(g) ⇄ 2HI(g)
K of equilibrium is:
K = [HI]² / [H2] [I2]
<em>Where [] are concentrations at equilibrium</em>
And Q is:
Q = [HI]² / [H2] [I2]
<em>Where [] are actual concentrations of the reactants.</em>
<em />
When the reaction is in equilibrium, K=Q.
But as [I2] is increased, Q decreases and Q was < K
The only concentration that increases is [I2], doing partial pressure of hydrogen decreased, iodine increased
