2 mol of Hg(g) is combusted in a fixed volume bomb calorimeter with excess of O2 at 298 K and 1 atm

Q: Why does the bomb calorimeter measure $\Delta U$ and not $\Delta H$?

Enthalpy ($\Delta H$) is the heat exchanged at constant pressure. A bomb calorimeter is a rigid, sealed steel container, meaning the volume is constant ($\Delta V = 0$). Therefore, no work is done ($P\Delta V = 0$), and all the heat exchanged corresponds directly to the change in internal energy ($\Delta U$).

Q: Why did we divide the total heat by 2?

The problem states that 2 moles of Hg(g) were combusted to produce the temperature rise. Standard thermodynamic values, like molar enthalpy, are always defined per 1 mole of the substance. So, we must divide the total energy by 2.

Q: How do we calculate $\Delta n_g$?

$\Delta n_g$ is the difference in the number of moles of gaseous products and gaseous reactants. For the reaction $\text{Hg(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{HgO(s)}$, the product is solid (0 gaseous moles), and the reactants have $1 + 0.5 = 1.5$ gaseous moles. Thus, $\Delta n_g = 0 - 1.5 = -1.5$.

Q: Why did we need the enthalpy of formation of Hg(g)?

The standard enthalpy of formation of a compound is the enthalpy change when 1 mole of it is formed from its elements in their standard states. At 298 K, the standard state of mercury is liquid, not gas. The reaction we analyzed was for Hg(g). We had to use the given enthalpy of formation of Hg(g) to 'convert' our gaseous mercury back to liquid mercury using Hess's Law.

Q: What if I forgot to convert R to kJ?

This is a classic trap! The internal energy $\Delta U$ is in kilojoules (kJ), but the universal gas constant R is given in joules (8.3 J/K·mol). If you don't divide R by 1000, your $\Delta n_g RT$ term will be 1000 times too large, leading to a completely wrong answer. Always check your units!

Question

2 mol of Hg(g) is combusted in a fixed volume bomb calorimeter with excess of O$_2$ at 298 K and 1 atm into HgO(s). During the reaction, temperature increases from 298.0 K to 312.8 K. If heat capacity of the bomb calorimeter and enthalpy of formation of Hg(g) are 20.00 kJ K$^{-1}$ and 61.32 kJ mol$^{-1}$ at 298 K, respectively, the calculated standard molar enthalpy of formation of HgO(s) at 298 K is X kJ mol$^{-1}$. The value of |X| is ______.
[Given : Gas constant R = 8.3 J K$^{-1}$ mol$^{-1}$]

Sigma Percentile · Accepted Answer

### The Bomb Calorimeter and the Hidden State of Mercury

Welcome to a beautiful and tricky problem from Thermodynamics! This question tests not only your ability to crunch numbers but also your conceptual clarity regarding standard states and the First Law of Thermodynamics. Let's break it down step-by-step.

### Analyzing the Setup

We are given a bomb calorimeter in which $2 	ext{ mol}$ of $	ext{Hg(g)}$ is combusted. A bomb calorimeter is a rigid, sealed steel container. Because it is rigid, the volume cannot change during the reaction. This means $\Delta V = 0$, and consequently, the work done ($P\Delta V$) is zero.

According to the First Law of Thermodynamics, $\Delta U = q + w$. Since $w = 0$, the heat exchanged at constant volume ($q_v$) is exactly equal to the change in internal energy ($\Delta U$).

### Calculating the Internal Energy Change

The heat released during the combustion is absorbed by the calorimeter, causing its temperature to rise. We can calculate this heat using the formula:

$$q = C \Delta T$$

Given the heat capacity $C = 20.00 	ext{ kJ K}^{-1}$ and the temperature change $\Delta T = 312.8 	ext{ K} - 298.0 	ext{ K} = 14.8 	ext{ K}$, the total heat released is:

$$|q| = 20.00 	imes 14.8 = 296 	ext{ kJ}$$

Since heat is released (exothermic reaction), the change in internal energy for the $2 	ext{ mol}$ of $	ext{Hg(g)}$ is $\Delta U_{	ext{2 mol}} = -296 	ext{ kJ}$.

However, standard thermodynamic properties are defined per mole. Therefore, for $1 	ext{ mol}$ of $	ext{Hg(g)}$, the internal energy change is:

$$\Delta U = \frac{-296}{2} = -148 	ext{ kJ mol}^{-1}$$

### The Master Equation: From $\Delta U$ to $\Delta H$

We need the enthalpy change ($\Delta H$) for the reaction. The relationship between $\Delta H$ and $\Delta U$ is given by:

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

Let's write the balanced chemical equation for the combustion of $1 	ext{ mol}$ of $	ext{Hg(g)}$:

$$	ext{Hg(g)} + \frac{1}{2}	ext{O}_2	ext{(g)} ightarrow 	ext{HgO(s)}$$

Here, $\Delta n_g$ is the change in the number of moles of gases. The product is a solid ($0$ gaseous moles), and the reactants have $1 + 0.5 = 1.5$ gaseous moles. Thus:

$$\Delta n_g = 0 - 1.5 = -1.5$$

Now, we substitute the values into our master equation. **Crucial Step:** Remember to divide $R$ ($8.3 	ext{ J K}^{-1}	ext{mol}^{-1}$) by $1000$ to convert it to $	ext{kJ}$!

$$\Delta H = -148 + (-1.5) 	imes \left(\frac{8.3}{1000}ight) 	imes 298$$

$$\Delta H = -148 - 3.7101 = -151.71 	ext{ kJ mol}^{-1}$$

### The Standard State Trap

Are we done? Not quite! The question asks for the **standard molar enthalpy of formation** of $	ext{HgO(s)}$. By definition, the standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its elements in their **standard states**.

At $298 	ext{ K}$, the standard state of mercury is **liquid**, not gas! The reaction we just analyzed was for gaseous mercury. We need the enthalpy change for:

$$	ext{Hg(l)} + \frac{1}{2}	ext{O}_2	ext{(g)} ightarrow 	ext{HgO(s)}$$

We are given the enthalpy of formation of $	ext{Hg(g)}$ as $61.32 	ext{ kJ mol}^{-1}$. This represents the vaporization of mercury:

$$	ext{Hg(l)} ightarrow 	ext{Hg(g)} \quad \Delta H = 61.32 	ext{ kJ mol}^{-1}$$

Using Hess's Law, we can add this vaporization reaction to our combustion reaction:

$1) \quad 	ext{Hg(g)} + \frac{1}{2}	ext{O}_2	ext{(g)} ightarrow 	ext{HgO(s)} \quad \Delta H_1 = -151.71 	ext{ kJ mol}^{-1}$

$2) \quad 	ext{Hg(l)} ightarrow 	ext{Hg(g)} \quad \Delta H_2 = 61.32 	ext{ kJ mol}^{-1}$

Adding these two equations cancels out $	ext{Hg(g)}$ and gives us the required formation reaction. The total enthalpy change is:

$$\Delta_f H^\circ(	ext{HgO}) = -151.71 + 61.32 = -90.39 	ext{ kJ mol}^{-1}$$

The magnitude $|X|$ is therefore **$90.39$**.

Enter Numerical Value:

Visualized Solution

\text{Visualizing the Bomb Calorimeter}

q_v = \Delta U

\text{Calculating Total Heat}

\Delta U_{\text{total}} = -296 \text{ kJ}

\Delta U_{\text{molar}} = -148 \text{ kJ mol}^{-1}

\Delta H = \Delta U + \Delta n_g RT

\text{Finding } \Delta n_g

\text{Substituting Values}

\Delta H_{\text{rxn}} = -151.71 \text{ kJ mol}^{-1}

\text{The Standard State Trap}

\text{Applying Hess's Law}

\Delta_f H^\circ(\text{HgO}) = -90.39 \text{ kJ mol}^{-1}

The Sigma Insight: First Law of Thermodynamics

The Bomb Calorimeter and the Hidden State of Mercury

Analyzing the Setup

Calculating the Internal Energy Change

The Master Equation

The Standard State Trap

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Visualizing the Bomb Calorimeter} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q_v = \Delta U .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \Delta U_{\text{total}} = -296 \text{ kJ} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \Delta H = \Delta U + \Delta n_g RT .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Finding } \Delta n_g .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Applying Hess's Law} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }