Which one of these is a gas NaNo3 KL KCL NH3
1 answer:
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Explanation:
For the given reaction:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
where x and y are order wrt to and
According to collision theory , the molecules must collide for a reaction to take place. According to collision theory , the rate of a reaction is proportional to rate of collision of reactants.
Thus with an increase in concentration of reactants , the rate of reaction also increases. This is because if the concentration of reactants increases , the chances of collision between molecules also increases and thus more products wil be formed which in turn increases the rate of reaction.
The reaction described above is the formation of an acetal. The initial starting material has a central carbonyl and two terminal alcohol functional groups. In the presence of acid, the carbonyl will become protonated, making the carbon of the carbonyl susceptible to nucleophilic attack from one of the alcohols. The alcohol substitutes onto the carbon of the carbonyl to provide us with the intermediate shown.
The intermediate will continue to react in the presence of acid and the -OH that was once the carbonyl will become protonated, turning it into a good leaving group. The protonated alcohol leaves and is substituted by the other terminal alcohol to give the final acetal product. The end result of the overall reaction is the loss of water from the original molecule to give the spiroacetal shown in the image provided.
The intermediate will continue to react in the presence of acid and the -OH that was once the carbonyl will become protonated, turning it into a good leaving group. The protonated alcohol leaves and is substituted by the other terminal alcohol to give the final acetal product. The end result of the overall reaction is the loss of water from the original molecule to give the spiroacetal shown in the image provided.