Consider the reaction, N2 + 3H2 - 2NH3 carried out at constant temperature and pressure. If ΔH and

Q: Why do we only consider gaseous moles for $\Delta n_g$?

Solids and liquids have negligible volume compared to gases. The work done by expansion or compression ($P\Delta V$) is almost entirely due to the change in the number of moles of gas. Therefore, we only count gaseous species.

Q: Is $\Delta E$ the same as $\Delta U$?

Yes, $\Delta E$ and $\Delta U$ both represent the change in internal energy of the system. Different textbooks and exam boards use different notations, but they mean exactly the same thing.

Q: What if the reaction was carried out at constant volume instead of constant pressure?

At constant volume, the heat exchanged is exactly equal to the change in internal energy ($\Delta E$). The relationship $\Delta H = \Delta E + V\Delta P$ would apply, but the standard $\Delta n_g RT$ form assumes constant pressure.

Q: Can $\Delta H$ ever be exactly equal to $\Delta E$?

Absolutely! If a reaction has the same number of moles of gaseous products as gaseous reactants (so $\Delta n_g = 0$), then the $P\Delta V$ work is zero, and $\Delta H = \Delta E$. An example is $\text{H}_2(g) + \text{I}_2(g) \longrightarrow 2\text{HI}(g)$.

Q: Does the temperature affect the inequality?

As long as the absolute temperature $T$ is greater than 0 K (which it always is in practical chemistry), the term $RT$ is positive. Thus, the sign of $\Delta n_g$ solely determines whether $\Delta H$ is greater than, less than, or equal to $\Delta E$.

Question

Consider the reaction,
$	ext{N}_2 + 3	ext{H}_2 \longrightarrow 2	ext{NH}_3$ carried out at constant temperature and pressure. If $\Delta H$ and $\Delta E$ are the enthalpy and internal energy changes for the reaction, which of the following expressions is true ?

Options:

(A) $\Delta H > \Delta E$

(B) $\Delta H < \Delta E$

(C) $\Delta H = \Delta E$

(D) $\Delta H = 0$

Sigma Percentile · Accepted Answer

### The Dance of Enthalpy and Internal Energy

In the realm of chemical thermodynamics, understanding how energy flows and transforms during a reaction is paramount. Let's dive into a classic problem involving the famous Haber process for the synthesis of ammonia. We are given the reaction:

$	ext{N}_2(g) + 3	ext{H}_2(g) \longrightarrow 2	ext{NH}_3(g)$

Our mission is to determine the relationship between the enthalpy change ($\Delta H$) and the internal energy change ($\Delta E$) for this process when carried out at constant temperature and pressure.

### The Master Equation

To connect enthalpy and internal energy, we rely on a fundamental thermodynamic relation derived from the First Law of Thermodynamics. For reactions involving ideal gases at constant temperature and pressure, the relationship is elegantly expressed as:

$\Delta H = \Delta E + \Delta n_g RT$

Here, $R$ is the universal gas constant, $T$ is the absolute temperature in Kelvin, and $\Delta n_g$ is the star of our show: the change in the number of moles of gaseous species.

### Decoding $\Delta n_g$

What exactly is $\Delta n_g$? It is simply the total number of moles of gaseous products minus the total number of moles of gaseous reactants.

**Crucial Trap:** We strictly only count the species that are in the gaseous state. Solids and liquids have negligible volume compared to gases, so their contribution to expansion work is practically zero.

Let's calculate $\Delta n_g$ for our specific reaction. On the reactant side, we have $1$ mole of nitrogen gas and $3$ moles of hydrogen gas, giving us a total of $4$ moles of gaseous reactants ($n_{	ext{reactants}} = 4$). On the product side, we have $2$ moles of ammonia gas ($n_{	ext{products}} = 2$).

Therefore, the change in gaseous moles is:

$\Delta n_g = 2 - 4 = -2$

### The Final Deduction

Now, let's substitute this value of $-2$ back into our master equation. We get:

$\Delta H = \Delta E + (-2)RT$

$\Delta H = \Delta E - 2RT$

Notice the negative sign here; it's the key to our final conclusion. Since the universal gas constant $R$ and the absolute temperature $T$ are always positive quantities, the term $2RT$ is inherently positive.

If we are subtracting a positive quantity ($2RT$) from the internal energy change ($\Delta E$) to obtain the enthalpy change ($\Delta H$), it mathematically dictates that $\Delta H$ must be strictly less than $\Delta E$.

**$\Delta H < \Delta E$**

And there we have it! By simply analyzing the stoichiometry of the gaseous species, we can confidently predict the relationship between the heat exchanged at constant pressure and the heat exchanged at constant volume.

Animated Solution for Chemistry - Chemical Thermodynamics: Consider the reaction, $N_{2} + 3 H_{2} ⟶ 2 NH_{3}$ carried out at constant temperature and pressure. If $Δ H$ and $Δ E$ are the enthalpy and internal energy changes for the reaction, which of the following expressions is true ?

Select Answer:

Visualized Solution

$The Reaction$

$First Law of Thermodynamics$

$Understanding Δ n_{g}$

$Calculating Δ n_{g}$

$Substituting Δ n_{g}$

$Final Conclusion$

$The Way Forward$

The Sigma Insight: First Law of Thermodynamics

The Dance of Enthalpy and Internal Energy

The Master Equation

Decoding $Δ n_{g}$

The Final Deduction

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Reaction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } First Law of Thermodynamics

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Understanding Δng​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Δng​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Δng​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Conclusion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward