Solving Rate of Reaction Problems: Ethyl Ethanoate and Sodium Hydroxide

[jsmith613]

Homework Statement

http://www.mrev.org.uk/A-Level_Site/New%20A-Level%20Unit%204/Rates.html
(the answers are at the top - see link)

Could someone please help me with the question (about half way through) on ethyl ethanoate and sodium hydroxide.
in cii) we identified the half life doubles
ciii) we identified the order is second order

for part civ) we are asked is this the order for OH- ions or total order.
I would have said OH- order BUT the answer is total order.
Please could someone explain this (the mark schemes explanation is ester and NaOH are equimolar at the start - how is this relevant)

thanks

Homework Equations

NONE

The Attempt at a Solution

I have shown my attempts above in the body of this question

 

[jsmith613]

questions like this come up often and I am worried it may come up in mine so i would really like to understand the theory behind it

 

[jsmith613]

i read a few books and they all agree the graph should only show order with respect to OH ions
why is this not the case here

 

[nobahar]

It might be better if you present the relevant information here. I couldn't find what you're talking about!

 

[DeeNos]

.

As a scientist, it is important to carefully analyze and interpret data to arrive at accurate conclusions. In this case, we are dealing with a rate of reaction problem involving ethyl ethanoate and sodium hydroxide. The given information suggests that the reaction follows a second-order kinetics, as evidenced by the doubling of the half-life in part cii). This means that the rate of the reaction is directly proportional to the concentration of both reactants, ethyl ethanoate and sodium hydroxide.

In part civ), we are asked to identify the order of the reaction in terms of OH- ions or the total order. The given answer states that the total order is second order, which may seem contradictory as we have previously identified the second order as being related to both reactants. However, upon further analysis, we can see that the total order is indeed second order because the concentration of OH- ions is directly proportional to the concentration of sodium hydroxide. In other words, as the concentration of OH- ions increases, the concentration of sodium hydroxide also increases, resulting in a second-order relationship.

The explanation given in the mark scheme about the initial equimolar concentrations of ethyl ethanoate and sodium hydroxide is relevant because it helps us understand the relationship between the concentrations of both reactants. In a second-order reaction, the rate of the reaction is affected by the concentrations of both reactants. In this case, since the initial concentrations of both reactants are equal, any changes in their concentrations will have an equal effect on the rate of the reaction. Therefore, the total order of the reaction is second order, taking into account both reactants.

In conclusion, it is important to carefully consider all relevant factors and data when solving rate of reaction problems. The given answer may seem counterintuitive at first, but upon further analysis, we can see that it is indeed correct and takes into account the equimolar concentrations of the reactants and their proportional relationship.