Enthelpy of a reaction question

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In the gas-phase reaction 2NO + O2 --- 2NO2 conducted in a constant-volume container at constant temperature, the measured heat change corresponds to a change in internal energy, not enthalpy. The equations indicate that under constant volume, the change in internal energy (delta U) equals the heat exchanged (q), while enthalpy (delta H) also equals q but includes pressure-volume work, which is zero in a constant-volume scenario. The confusion arises from the constant temperature condition, where heat can be transferred out of the system without changing the temperature inside the vessel. In this case, the internal energy change is directly related to the heat transfer, while enthalpy remains equal to internal energy due to the lack of volume change. Thus, both quantities are equal, but the focus remains on internal energy due to the reaction conditions.
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Suppose that the gas-phase reaction 2NO + O2 --- 2NO2 were carried out in a constant-volume container at constant temperature. Would the measured heat change represent a change in enthalpy or a change in internal energy? If there is a difference, which quantity is larger for this reaction? Explain.delta U = q + w
delta H = delta U + P(delta V)I've been able to reduce the two equations so that delta U = q given a constant-pressure and volume container. Of course, delta H is also equal to q. So from that, the internal energy and enthalpy change would be equal. Or so I would assume.

What is throwing me off is the constant-temperature. If the value of q is changing but the temperature is constant and so is the ability to work... I just don't understand how the temperature cannot change given the other conditions if a change in heat occurs.

Any help?
 
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Heat is transferred outside, so that temperature inside of the vessel doesn't change.
 

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