Solving Rate of Reaction Problems: Ethyl Ethanoate and Sodium Hydroxide

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The discussion focuses on a homework problem involving the reaction between ethyl ethanoate and sodium hydroxide, specifically addressing the reaction order. Participants clarify that the half-life of the reaction doubles, indicating it is second order. A key point of confusion arises regarding whether the order refers to the hydroxide ions (OH-) or the total order of the reaction. The mark scheme states it is the total order due to the initial equimolar concentration of both reactants, which is crucial for understanding the overall reaction kinetics. This highlights the importance of considering both reactants in determining the reaction order.
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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
 
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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
 
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
 
It might be better if you present the relevant information here. I couldn't find what you're talking about!
 
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