Enthelpy change of a neutralisation reaction

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Discussion Overview

The discussion revolves around calculating the enthalpy change of a neutralization reaction involving potassium carbonate and hydrochloric acid. Participants explore the application of thermodynamic equations and the specifics of the reaction conditions, including temperature changes and mass measurements.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents the reaction and attempts to apply the equation ΔH = -qmΔT, seeking clarification on the variable 'm' in the context of the reaction.
  • Another participant questions what undergoes the temperature change of 6°C, indicating a need for clarity in the experimental setup.
  • Some participants assert that the reaction is exothermic and emphasize that only specific types of enthalpy changes, such as neutralization, combustion, and formation, are typically considered in this context.
  • A later reply challenges the understanding of the initial question, suggesting that the application of formulas requires a proper understanding of what they describe.

Areas of Agreement / Disagreement

Participants express differing views on the application of the enthalpy change formula and the specifics of the temperature change, indicating that the discussion remains unresolved with multiple competing interpretations.

Contextual Notes

There are limitations regarding the clarity of the experimental conditions and the definitions of variables used in the equations, which remain unresolved in the discussion.

IDK10
Messages
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Homework Statement


Given the reaction K2CO3(s) + 2HCl(aq) → 2KCl(aq) + CO2(g) + H2O(l), I need to find the enthalpy change in J, given that:
The specific heat capacity of water is 4.2J g-1 °C-1, and
The mass change of potassium carbonate is 2.98g, and the temperature change was °6C, and the volume of acid used was 30cm3.

Homework Equations


ΔH = -qmΔT
number of moles of acid neutralised = number of moles of water formed

The Attempt at a Solution


Considering it is a neutralisation reaction, I did the ΔH = -qmΔT equation. So, -4.2x-2xm=x
x/(2x0.03)
But what do I use for m?
 
Physics news on Phys.org
What undergoes a temperature change of 6°C?
 
mjc123 said:
What undergoes a temperature change of 6°C?
Exothermic

We only do neutralisation, combustion and formation enthalpy changes.
 
IDK10 said:
Exothermic

We only do neutralisation, combustion and formation enthalpy changes.

I am afraid you have not understood nor answered the mjc's question, which points you in the right direction.

You can't blindly apply any formula without knowing what it describes.
 

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