Enthelpy at constant pressure.

In summary, the textbook derived the equation ΔH = ΔE + ΔnRT for enthalpy from the fact that ΔH = ΔE + PΔV at constant pressure. However, this derivation assumes that temperature is constant, which may not always be the case. The text also provides an example of an exothermic combustion reaction with a negative heat value, but the equation used assumes constant temperature. This assumption is based on the idea that the heat given off will usually increase the temperature of the surroundings. Additionally, for an ideal gas, the equation would be ΔH = ΔE + Δ(nRT) if the temperature changes. Overall, the heat given off in a reaction does not necessarily affect the temperature of the substance
  • #1
AcidRainLiTE
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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.
 
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  • #2
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)
 
  • #3
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?
 
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Related to Enthelpy at constant pressure.

What is Enthalpy at Constant Pressure?

Enthalpy at constant pressure is a thermodynamic property that describes the heat content of a system at a constant pressure. It takes into account both the internal energy of a system and the work done by or on the system.

How is Enthalpy at Constant Pressure Calculated?

Enthalpy at constant pressure can be calculated using the equation H = U + PV, where H is enthalpy, U is internal energy, P is pressure, and V is volume. This equation is typically used for ideal gas systems.

What is the Significance of Enthalpy at Constant Pressure?

Enthalpy at constant pressure is significant because it allows us to understand the energy transfer that occurs during a process at a constant pressure. It is often used in chemical and physical processes to determine the heat exchanged between a system and its surroundings.

How Does Enthalpy at Constant Pressure Differ from Enthalpy at Constant Volume?

Enthalpy at constant pressure and enthalpy at constant volume are two different thermodynamic properties. Enthalpy at constant pressure includes the work done by or on the system, while enthalpy at constant volume does not. Additionally, enthalpy at constant pressure is typically used for open systems, while enthalpy at constant volume is used for closed systems.

What are Some Real-World Applications of Enthalpy at Constant Pressure?

Enthalpy at constant pressure has many real-world applications, including in chemical reactions, power plants, and refrigeration systems. It is also used in industries such as food processing and pharmaceuticals to control and optimize heating and cooling processes.

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