At 25°C, 50 g of iron reacts with HCl to form FeCl2. The evolved hydrogen gas expands against a cons

Q: Why do we use $\Delta n_g RT$ instead of $p\Delta V$?

Because we don't know the initial and final volumes of the system. However, we can easily calculate the moles of gas produced and use the ideal gas assumption to substitute $p\Delta V$ with $\Delta n_g RT$.

Q: Why is the work done positive when the formula has a negative sign?

The formula $W = -p_{\text{ext}}\Delta V$ calculates the work done *on* the gas. When a gas expands, it does work *on* the surroundings, so the work done *by* the gas is the negative of $W$, making it a positive value.

Q: Do we need to consider the volume of the solid iron and liquid HCl?

No, the volume occupied by solids and liquids is extremely small compared to the volume of the gas produced. Therefore, their contribution to the total volume change $\Delta V$ is negligible and can be safely ignored.

Q: Why is the temperature taken as 298 K?

In thermodynamics equations involving the ideal gas constant $R$, temperature must always be in Kelvin. The problem states the reaction happens at $25^{\circ}\text{C}$, which converts to $25 + 273 = 298 \text{ K}$.

Q: What if the pressure wasn't constant?

If the external pressure varied during the expansion, we couldn't use the simple formula $W = -p_{\text{ext}}\Delta V$. Instead, we would need to integrate the pressure with respect to volume, $W = -\int p \, dV$, which requires knowing exactly how pressure changes with volume.

Question

At $25^{\circ}	ext{C}$, $50	ext{ g}$ of iron reacts with $	ext{HCl}$ to form $	ext{FeCl}_2$. The evolved hydrogen gas expands against a constant pressure of $1	ext{ bar}$. The work done by the gas during this expansion is ...... $	ext{J}$. (Round off to the nearest integer)
[Given, $R = 8.314	ext{ J mol}^{-1}	ext{ K}^{-1}$. Assume, hydrogen is an ideal gas]
[Atomic mass off Fe is $55.85	ext{ u}$]

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are observing a chemical reaction in a beaker fitted with a movable piston. Inside, solid iron is reacting vigorously with aqueous hydrochloric acid. As the reaction proceeds, bubbles of hydrogen gas start to form and rise.

This newly formed gas needs space, so it pushes against the piston, expanding against the constant atmospheric pressure of the room. In thermodynamics, when a system pushes against its surroundings, it does work. Our goal is to calculate exactly how much work this expanding hydrogen gas performs.

### The Master Equation

To find the work done, we first need to know exactly what is happening chemically. The balanced chemical equation for this reaction is:

$$ 	ext{Fe}(s) + 2	ext{HCl}(aq) \longrightarrow 	ext{FeCl}_2(aq) + 	ext{H}_2(g) $$

Notice that for every mole of solid iron consumed, exactly one mole of hydrogen gas is produced.

Now, how do we calculate the work? The formula for work done by a gas expanding against a constant external pressure is:

$$ W = -p_{	ext{ext}}\Delta V $$

However, we don't know the initial and final volumes. But we do know that the hydrogen gas behaves ideally. Using the ideal gas law, $pV = nRT$, we can rewrite the work equation in terms of the change in the number of moles of gas:

$$ W = -\Delta n_g RT $$

Here, $\Delta n_g$ is the difference between the moles of gaseous products and gaseous reactants.

### Calculating the Moles

Let's figure out how much hydrogen gas we actually have. We are given $50 	ext{ g}$ of iron. To find the moles of iron, we divide its mass by its atomic mass:

$$ n_{	ext{Fe}} = \frac{50}{55.85} \approx 0.8952 	ext{ mol} $$

Since the stoichiometry of the reaction tells us that $1 	ext{ mole}$ of iron produces $1 	ext{ mole}$ of hydrogen gas, the moles of hydrogen gas produced is also $0.8952 	ext{ mol}$.

Because there are no gaseous reactants, the change in gaseous moles is simply the moles of hydrogen produced:

$$ \Delta n_g = 0.8952 - 0 = 0.8952 	ext{ mol} $$

### Final Calculation

Now, we have all the pieces of the puzzle. Let's substitute them into our master equation. The universal gas constant $R$ is $8.314 	ext{ J mol}^{-1} 	ext{ K}^{-1}$, and the temperature is $25^{\circ}	ext{C}$, which is $298 	ext{ K}$.

$$ W = -0.8952 	imes 8.314 	imes 298 $$

$$ W = -2217.92 	ext{ J} $$

The negative sign indicates that work is done *by* the system (the gas) on the surroundings. The question asks for the magnitude of the work done by the gas, which is the positive value.

Rounding off to the nearest integer, we get our final answer:

**2218 J**

Enter Numerical Value:

Visualized Solution

Visualizing the Expansion of $H_{2}$ Gas

Formula for Work Done $W$

Calculating Moles $Δ n_{g}$

Substituting Values into $W = - Δ n_{g} R T$

Final Calculation of Work Done

What if $Δ V = 0$ ?

The Sigma Insight: First Law of Thermodynamics

Analyzing the Setup

The Master Equation

Calculating the Moles

Final Calculation

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Visualized Solution

Visualizing the Expansion of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H2​ Gas

Formula for Work Done .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } W

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

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