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I would like to ask two separate questions. But first, so that we are on the same page, the experiment used a constant-pressure calorimeter, and it is a neutralization reaction of 1M of HCl and 1M of NaOH.

1. What is the difference between the units of ΔH: kJ and kJ mol^-1?

As I have read in Socratic at the end of the page, there are many units to represent the change in enthalpy. However, we are given a theoretical enthalpy of -2.79 kJ but my teacher gave an equation of Q_calorimeter/n_H2O thus my answer is -1.16 kJ mol^-1. How am I able to evaluate this by getting the percentage error? Should I multiply my answer to something with "mol^-1"? If so which mole would it be, the whole reaction or each species?

2. If you are not given data on the initial temperature of the calorimeter, is it correct to assume that its initial temperature is the initial temperature of the substance that is initially poured in (i.e. NaOH)?

In the other part of the activity, the initial temperature of the cold water is explicitly labeled also as the initial temperature of the calorimeter, and the cold water is initially poured into the calorimeter. Inferring from this, is my conclusion correct?

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    $\begingroup$ It is long time but the calorimeter equivalent makes sense as an absolute value to me. Perhaps I am wrong but it should be how much energy it takes. The calorimeter is that one calorimeter. $\endgroup$ – Alchimista Sep 11 at 17:17
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    $\begingroup$ This is how I would proceed: 2) ask your teacher or schoolmates about such details. You are presumably allowed to take some liberties with respect to accuracy. 1) use standard propagation of errors (look this up if you are not familiar) to compute the uncertainty in the ratio heat/amount of water formed. As the reaction goes to completion the question is which reagent is in excess. You need to estimate uncertainties in the amount of the limiting reagent (either NaOH or HCl). I you don't know which is limiting assume they are equal and estimate the uncertainty in either one. $\endgroup$ – Buck Thorn Sep 12 at 8:01

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