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Is the reaction of silver nitrate and copper endothermic or exothermic?
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Michael Ahler
Is the reaction of silver nitrate and copper endothermic or exothermic?
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.
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.
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 …More
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.
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.
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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.
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.
More
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