The reaction with HBr is a nucleofilic adition with the following steps:
Step 1: Carbocation formation
Step 2: Product formation (addition of Br-)
In the first step, the more sustituted the carbocation is the more stable. Meaning this that the most stable will be a terciary C (such as in ter buthene), the less stable the primary C (ethene for example) and between them, the secondary C (propene).
Analizing that and knowing that this mechanism of reaction follows the rule of Markovnikov, you can determine the reaction speed of an alkene.
Also the more resonance intermediates it has, the slower the reacrion is.
The first step in the reaction is the double bond of the Alkene going after the H of HBr. This protonates the Alkene via Markovnikov's rule, and forms a carbocation. The stability of this carbocation dictates the rate of the reaction.
<span>So to solve your problem, protonate all your Alkenes following Markovnikov's rule, and then compare the relative stability of your resulting carbocations. Tertiary is more stable than secondary, so an Alkene that produces a tertiary carbocation reacts faster than an Alkene that produces a secondary carbocation.
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The equation that correctly represent the reaction for formation of ammonia is
N2+ 3H2 → 2NH3 (answer D)
1 mole of nitrogen gas (N2) react with 3 moles of hydrogen gas (H2) to form 2 moles of ammonia ( NH3). This is in a process known as haber process were iron is used as a catalyst and reaction take place that a higher temperature and pressure. The process is exothermic hence energy is released.
