# Enthalpy of a reaction question

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?

## Answers and Replies

Borek
Mentor
Heat is transferred outside, so that temperature inside of the vessel doesn't change.