1
$\begingroup$

When a chemical reaction dissipate heat to the surrounding, the molecules that participate in the chemical reaction will have high kinetic energy and will transmit some of the kinetic energy to the surrounding and the remaining kinetic energy will increase their temperature.

When we say that heat is dissipated to the surrounding we are talking about "some of the kinetic energy" right? Also when we write that heat dissipated is equal to $\Delta H$, it is included the remaining kinetic energy?

Can somebody confirm whether my understanding is correct or clarify where I went wrong if it is not?

Sorry, but I am far from chemistry field.

$\endgroup$
  • 1
    $\begingroup$ By DH do you mean $\Delta H$?. If that is the cause, the answer is not always. Heat dissipated is known as q (heat transfer) which is only equal to $\Delta H$ when pressure is constant. $\endgroup$ – Nanoputian Nov 17 '15 at 20:25
  • 2
    $\begingroup$ To expand on Nanoputian's answer, $-\Delta H$ is the amount that has to be removed (dissipated) to hold the temperature constant at its initial value. $\endgroup$ – Chet Miller Nov 17 '15 at 21:31
0
$\begingroup$

Let establish some thermochemistry terminology first:

Work: the total energy transferred by a system to another. There are several types of work. In thermodynamics, the most common one is pressure volume work, given by: $w = -p_{ext}\Delta V$

Internal Energy: the total kinetic and potential energy of a system

Enthalpy ($H$): is the total energy of a system. It also can be thought as the amount of energy that is required to make all the bonds in the system plus all the energy that is required to make room for the system. Hence it should be clear that it is given by the formula: $\Delta H = \Delta U + \Delta {(pV)}$

Heat Transfer (q): the amount of heat energy (kinetic energy) that is transferred from a system given by the formula: $q = \Delta U -w = \Delta U + p_{ext}\Delta V ~~$ (when only pressure volume work is done)

In your question, you are actually referring to $q$ (heat transfer), not $\Delta H$. This is because $\Delta H$ also accounts for the change in pressure and volume.

However under certain conditions, $q$ and $\Delta H$ can be equal. If the external pressure is constant (which is a common case as most experiments are done in the open atmosphere which has a constant pressure), enthalpy becomes: $$\Delta H = \Delta U + p_{ext}\Delta V = q$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.