Is Temperature Constant in Exothermic Reactions?

Click For Summary
SUMMARY

The discussion centers on the relationship between temperature and enthalpy in exothermic reactions, specifically addressing the equation ΔH = ΔE + ΔnRT. The derivation of this equation assumes constant temperature, which raises questions when applied to exothermic reactions like combustion, where heat is released (q = -3264 kJ/mol). It is clarified that the heat released increases the temperature of the surroundings rather than lowering the temperature of the reacting substance, indicating that the substance's temperature can remain effectively constant during the reaction.

PREREQUISITES
  • Understanding of thermodynamics, specifically enthalpy and internal energy.
  • Familiarity with the ideal gas law and its implications in chemical reactions.
  • Knowledge of exothermic reactions and heat transfer principles.
  • Basic grasp of chemical kinetics and reaction mechanisms.
NEXT STEPS
  • Study the derivation and implications of the equation ΔH = ΔE + ΔnRT in detail.
  • Research the concept of heat transfer in exothermic reactions and its effects on surroundings.
  • Explore the behavior of ideal gases under varying temperature and pressure conditions.
  • Investigate real-world applications of enthalpy changes in combustion reactions.
USEFUL FOR

Chemistry students, chemical engineers, and professionals involved in thermodynamics and reaction engineering will benefit from this discussion.

AcidRainLiTE
Messages
89
Reaction score
2
My textbook derived the equation ΔH = ΔE + ΔnRT (H is enthalpy, E is internal energy (heat exchanged + work )) from the fact that ΔH = ΔE + PΔV at constant pressure. This derivation, however, requires that temperature be constant; otherwise, the resulting equation would be ΔH = ΔE + ΔnRΔT (ΔT instead of T). A paragraph or two later they give an example of an exothermic combustion reaction where q = -3264 kJ/mol (where q is the heat absorbed by the reaction, since it is negative heat is given off). Then they go on to use the equation ΔH = ΔE + ΔnRT. But this equation assumes that temperature is constant. How can the temperature be constant when the heat of reaction is -3264 kJ/mol? Isn't heat given off thereby lowering the temperature of the combusted substance?

Thanks.
 
Physics news on Phys.org
Good question.

The text assumes that the temperature doesn't change, or changes negligibly. In fact, the heat given off will usually increase the temperature of the surroundings, and thus the substance.

Also, if the temperature changes, for an ideal gas, you'll have ΔH = ΔE + Δ(nRT)
 
Ok, so the heat given off (q) isn't referring to heat transferred and lost by the substance in question; rather, it is referring to heat created (by chemical bonds breaking and such) and given off in the reaction. So the actual heat of the substance could theoretically stay the same, since the heat given off is not taken from the subtance but, rather from the reaction. Am I understanding that correctly?
 
Last edited:

Similar threads

  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
14K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K