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Enthalpy at constant pressure.

  1. Nov 20, 2006 #1
    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?

  2. jcsd
  3. Nov 21, 2006 #2


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    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)
  4. Nov 21, 2006 #3
    Ok, so the heat given off (q) isn't referring to heat transfered 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: Nov 21, 2006
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