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Heat is required to make the reaction occur.

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11y ago
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Dilim

Lvl 1
2y ago
Correct!
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AnswerBot

2w ago

No, an exothermic reaction releases heat energy to the surroundings, resulting in a negative change in enthalpy (ΔH < 0). A positive change in enthalpy (ΔH > 0) indicates an endothermic reaction, which absorbs heat energy from the surroundings.

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15y ago

This is actually trickier than it sounds, because it depends on the particular conventions you're using. I studied thermodynamics in two different classes, one chemistry and and one physics, at the same time ... and the two textbooks used different conventions for the sign of delta-H. The physics book used a positive delta-H to mean that you could get work OUT of the system (exothermic), the chemistry book used a positive delta-H to mean you had to put work INTO the system (endothermic). The best thing is to review your textbook and see which convention they use.

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8y ago

Typically, exothermic reactions have negative enthalpy, But this may not always necessarily be the case.

To prove this we say that H=U+PV and U=Q+W (H=enthalpy, U=internal energy, P=pressure, V=volume, Q=heat, W=work).

H=U+PV

Substitute for U (see above)

H=Q+W+PV

Take the differential of dH:

dH=dQ+dW+d(PV)

Plug in for dW with dW=-P(deltaV)) (this is true for a ideal gas expanding against constant external pressure)

dH=dQ-P(deltaV)+d(PV)

Take the derivative of d(PV) using the chain rule

dH=dQ-P(deltaV)+P(deltaV)+V(deltaP)

Cancel like terms

dH=dQ+V(deltaP)

At constant pressure (which is what is normally observed in almost all naturally occurring environments the pressure is constant so the V(deltaP) term is 0 (change in volume is zero) and therefore dH=dQ. So in this case, if Q is negative so is enthalpy. However, if are at constant volume conditions (example: a bomb calormiter) where the pressure can change it may be possible to have a V(deltaP) term that is greater than the negative Q (indicating the reaction is exothermic while still giving you a positive enthalpy.

Example: dQ=-1000J (Exothermic process) V=(5L) deltaP=2atm

dH=dQ+V(deltaP)

dH=-1000J+10L*atm=-1000J+1013.25J=13.25... (10Latm=1013.25J)

dH is positive while the process is exothermic. I understand these conditions are extreme and unlikely, but they are being used to illustrate a point - mathematically, and potentially IRL, it is plausible to have an exothermic process with a positive change in enthalpy.

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13y ago

yes

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12y ago

negative

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