# Is the reaction of silver nitrate and copper endothermic or exothermic?

Is the following reaction endothermic or exothermic and why?

$$\ce{2 AgNO3 + Cu -> Cu(NO3)2 + 2 Ag}$$

EDIT: I have a feeling that it is exothermic due to the fact that it is spontaneous and will end in around 30 minutes(which I am guessing is because it will run out of energy to burn) and no source of energy is there for it to gain energy from, but is this right? Online, I found that I was right about it being exothermic but my real question is why? Is the fact that it is spontaneous the only means of support for this question?

• Hi and welcome to chemistry.stackexchange.com. Feel free to take a tour of the site. I improved the formatting of your post by adding MathJax. For more information on how to do so yourself, check out the help center, this meta-post or this one. Note that your question can be considered homework, and as per our homework policy, we require you to show your thoughts and work to prevent it getting closed. Please edit the post accordingly. – Jan Jun 5 '16 at 20:06
• You're basically saying that it's exothermic because it happens which of course isn't true. – Mithoron Jun 5 '16 at 20:21
• Many endothermic reactions are spontaneous as well. – Ivan Neretin Jun 5 '16 at 20:22

You may already know that the world is not about enthalpy but about Gibbs free energy. A reaction is sponanteous if the change Gibbs free energy is negative.

$$\Delta G = \Delta H - T \cdot \Delta S$$

As you see, enthalpy and entropy play into this. Looking at your reaction, we note that we are taking two ions out of a solution to release one into solution. Compare the following, slightly different description:

$$\ce{2 Ag+ (aq) + Cu (s) <=>> 2 Ag (s) + Cu^2+ (aq)}$$

From an entropic point of view, this is bad. Rather than having two dissolved ions, we have one on the products’ side. Therefore, we can assume $\Delta S < 0$, therefore $T \cdot \Delta S < 0$. Negating it gives a positive value.

But the reaction is spontaneous, so $\Delta G < 0$. This can only be true if $\Delta H$ is sufficiently negative to counteract the loss of entropy. Thus we conclude that $\Delta H < 0$ and the reaction is exothermic.

I have carried out the experiment myself and from observation, the reaction is exothermic (releases heat) even though I didn't quantitatively determine the value of heat change occurred.

• If you didn’t quantitatively determine the heat released, how can you be sure that the reaction is exothermic? As it stands, this answer needs additional references … – Jan Jun 6 '16 at 0:29