Enthelpy change of a neutralisation reaction

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SUMMARY

The discussion centers on calculating the enthalpy change (ΔH) for the neutralization reaction between potassium carbonate (K2CO3) and hydrochloric acid (HCl). The relevant equation used is ΔH = -qmΔT, where the specific heat capacity of water is 4.2 J g-1 °C-1, the mass of K2CO3 is 2.98 g, and the temperature change is 6°C. Participants emphasize the importance of understanding the context of the formula and the exothermic nature of the reaction, clarifying that the number of moles of acid neutralized equals the number of moles of water formed.

PREREQUISITES
  • Understanding of enthalpy changes in chemical reactions
  • Familiarity with the equation ΔH = -qmΔT
  • Knowledge of specific heat capacity and its application
  • Basic concepts of neutralization reactions
NEXT STEPS
  • Research the calculation of enthalpy changes for various types of reactions
  • Learn about the concept of exothermic and endothermic reactions
  • Explore the role of moles in chemical reactions and stoichiometry
  • Study the specific heat capacity of different substances and its implications in thermodynamics
USEFUL FOR

Chemistry students, educators, and anyone involved in thermodynamics or chemical reaction analysis will benefit from this discussion.

IDK10
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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?
 
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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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