Enthelpy of a reaction question

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SUMMARY

The gas-phase reaction 2NO + O2 --- 2NO2 conducted in a constant-volume container at constant temperature results in equal changes in enthalpy (ΔH) and internal energy (ΔU). The relationship ΔU = q + w and ΔH = ΔU + P(ΔV) indicates that under constant volume, where ΔV equals zero, ΔH simplifies to ΔU. Therefore, for this reaction, both ΔH and ΔU are equivalent to the heat transfer (q) occurring in the system, despite the constant temperature condition, which allows for heat transfer without temperature change.

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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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