Chemistry - Can the change in internal energy be nonzero if temperature is constant?
Solution 1:
TL;DR: Do not just memorise thermodynamics equations! And if you have an issue with the equations $\Delta U = 0$ or $\Delta H = 0$ for an isothermal process, read on...
The first problem
You said that an exothermic reaction corresponds to $\Delta U < 0$. This is not true. It is defined by $\Delta H < 0$.
However, we should still discuss this issue because $H$ is also a state function and in some cases $H$ is dependent only on temperature.
How to study thermodynamics
This kind of issue in thermodynamics frequently crops up here, and it is a very common mistake amongst students to indiscriminately use equations that they have learnt.
So, you have an equation that says $\Delta U = 0$ for an isothermal process, i.e. one at constant $T$. The important question here is not "what is the equation" or "what is the answer"! Instead, you should be asking yourself "how do I derive this result", and the answer will naturally follow. This should really apply to everything you do - how can you expect to apply a formula that you do not actually understand?
Furthermore, if you are able to actually understand how an equation comes about, then you don't need to memorise it - you can derive it. For example, what's the point of memorising $w = -nRT\ln(V_2/V_1)$, along with the conditions, if you can simply obtain the equation from the very definition of work $đw = -\int p \,\mathrm{d}V$ by substituting in $p = nRT/V$ and integrating? What's more, the fact that you actually substituted in the ideal gas law should tell you something about the conditions that accompany the equations - for one, it's only applicable to ideal gases. What about when you reach the equation $\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V$ and happily memorise it, only to find out later that there are actually three more equations that look exactly like it? Are you going to memorise all four? No - you need to know where they come from.
Internal energy
In general, the internal energy $U$ of a substance is a function of at least two variables, for example $T$ and $p$. However, it can be shown using statistical mechanics that for an ideal gas,
$$U = n\left(\frac{3NRT}{2} + U_0\right)$$
Here, $N$ is the number of atoms in one molecule of the gas (for example, $N = 2$ for $\ce{H2}$, and $N = 3$ for $\ce{H2O}$). $U_0$ is the molar internal energy at absolute zero (this contains terms such as electronic energy), and crucially, this depends on the exact identity of the gas. This result is one example of the equipartition theorem in action (although the theorem itself is more powerful).
We do not need to concern ourselves with its derivation now. The point is that: this equation can only be applied to a fixed amount of a fixed ideal gas at a time. For example: if I have one mole of $\ce{H2}$ at $298\mathrm{~K}$, it is going to have a different internal energy from one mole of $\ce{Ar}$ at $298\mathrm{~K}$, even though the amount of substance and the temperature are the same.
What does this mean?
Firstly, it means that for anything that is not an ideal gas, you cannot assume that $U = U(T)$ and therefore you cannot assume that $\Delta T = 0$ implies $\Delta U = 0$. Let's say I compress $10\mathrm{~g}$ of water by reversibly increasing the pressure from $1\mathrm{~atm}$ to $10\mathrm{~atm}$, at a constant temperature of $323\mathrm{~K}$. Is the change in internal energy equal to zero? No, because water is not an ideal gas.
Secondly, it means that if the amount of ideal gas changes in any way, $U$ is going to change. Let's say I have a balloon filled with $1\mathrm{~mol}$ of argon gas, and I add a further $2\mathrm{~mol}$ argon gas into the balloon, all at a fixed temperature of $298\mathrm{~K}$. Is the change in internal energy equal to zero? No, in fact it will triple because you are changing $n$ from $1\mathrm{~mol}$ to $3\mathrm{~mol}$!
Lastly, it means that if the chemical composition is changing in any way, you are automatically not allowed to say that $\Delta T = 0$ implies $\Delta U = 0$. And this is the case even if all reactants and products are ideal gases. Why not? Let's look at this reaction, and let's assume that everything there behaves as an ideal gas, and let's say that the reaction vessel is kept at a constant temperature of $300\mathrm{~K}$.
$$\ce{H2 (g) + Cl2 (g) -> 2HCl (g)}$$
Let's write an expression for the internal energy of the reactants. On the left-hand side, we have:
$$U_\mathrm{reactants} = \left[\frac{3}{2}RT + U_0(\ce{H2})\right] + \left[\frac{3}{2}RT + U_0(\ce{Cl2})\right]$$
On the right-hand side, we have:
$$U_\mathrm{products} = 3RT + 2U_0(\ce{HCl})$$
So, is $\Delta U$ equal to zero? No; even though the multiples of $RT$ cancel out with each other, the values of $U_0$ for the reacting species are different, which makes
$$\Delta U = 2U_0(\ce{HCl}) - U_0(\ce{H2}) - U_0(\ce{Cl2}) \neq 0.$$
Enthalpy
Let's go back to our (closed) system where there is only one ideal gas and it's not undergoing any chemical reaction. Within this system, we can safely say that if $\Delta T = 0$, then $\Delta U = 0$. So:
$$\begin{align} \Delta H &= \Delta (U + pV) \\ &= \Delta U + \Delta (pV) \\ &= \Delta (pV) \qquad \\ &= \Delta (nRT) \\ &= nR \Delta T \\ &= 0 \end{align}$$
We haven't made any assumptions here apart from the ideality of the gas (the fact that $n$ is constant came from the fact that it is a closed system). So, for an ideal gas, $\Delta T = 0$ implies $\Delta H = 0$.
However, we did use the fact that $\Delta U = 0$. Therefore, what I said about the cases above where $\Delta U \neq 0$ also applies here.
Regarding the terms endothermic and exothermic, they are used to describe chemical reactions. And for chemical reactions, there is absolutely no way you can say that constant temperature implies $\Delta U = 0$ or $\Delta H = 0$. That's why we can actually use the terms endothermic and exothermic. If every single chemical reaction had $\Delta H = 0$, life would be rather plain!
Solution 2:
Internal Energy is a function of temperature only in ideal gases, but for real life situation it is a function of temperature and even volume of the thermodynamic system, for example when solid water (ice) melts to liquid water then during melting the temperature of the "water system" is constant yet its internal energy of system increases, because it takes heat from surrounding so its internal energy will increase but keep a thermometer in contact with the water, you will see no change in temperature. This internal energy is now added to the KE of the molecules of water and to break the Hydrogen bonds b/w water molecules. (internal Energy of system = total KE of system + Total PE due to interaction b/w constituents of system), whereas in ideal gas there is no type of interaction b/w gas particles so no PE in ideal gas but only KE of ideal gas contributes to its internal energy, and as KE of an ideal gas is a function of temperature of ideal gas, hence for ideal gas internal energy depends only on temperature.
For your question enthalpy change is -ve for exothermic reactions not internal energy change, so in your book it was a misprint, btw in chemical reaction we assume the gas evolved to be ideal, then only can we say const temperature gives const U of system, but you know the world isn't always so ideal (real life state change and volume change considerations can completely change the ideal picture, so for the time being better not venture there and first get basic definitions and conditions for an equation correct).
Solution 3:
For exothermic reactions and endothermic reactions...as the reaction occurs, heat is evolved and absorbed by the reaction system respectively. Which means that there is a total change in the heat energy of the system and the change is suffered by the internal energy of the system. Let us consider the example: C+O2--->CO2...Now as this reaction occurs, the [externally] temperature remains a constant...that is we don't add heat nor do we take heat purposely from the system but then as the reaction proceeds and as it's an exothermic one, heat energy is evolved in the system which is not the problem of the constant temperature that was preset for the system...instead as the reaction proceeds, whatever energy (released by the breaking and creation of new bonds in the resultant compound) is indeed added to the total internal energy of the system.
Therefore, [delta] U can be non zero especially when the system id=s undergoing a reaction.
Hope you could understand that...?! Please let me know.