The enthalpy changes for the following processes are listed below
Cl2(g) = 2Cl(g), 242.3 kJ mol^-1
I

Q: Why do we multiply the bond dissociation energies by 1/2?

The standard enthalpy of formation is defined for exactly *one mole* of the product. To make 1 mole of ICl, we only need 0.5 moles of I2 and 0.5 moles of Cl2. Therefore, we only need half the energy to break those bonds.

Q: Why is the enthalpy for the ICl step negative? The question says it's 211.3!

The value 211.3 kJ/mol is the bond *dissociation* energy (breaking the bond), which requires energy (+). In our final step, we are *forming* the ICl bond from free atoms. Bond formation releases energy, so we must flip the sign to negative (-211.3 kJ/mol).

Q: What does "standard state" mean in this context?

The standard state is the most stable physical form of an element at 1 bar pressure and a specified temperature (usually 298 K). For chlorine, it's a diatomic gas (Cl2). For iodine, it's a diatomic solid (I2).

Q: Why isn't the final answer exactly 16.8?

The exact calculation yields 16.73 kJ/mol. In competitive exams, options are sometimes rounded to one decimal place. 16.73 rounds up to 16.7, but 16.8 is the closest provided option, likely due to slight variations in accepted standard values used by the examiner.

Q: Can I just subtract the reactant bond energies from the product bond energies?

Not directly! That shortcut only works if *all* reactants and products are in the gaseous state. Because iodine starts as a solid, you must account for the extra energy required to sublimate it into a gas first.

Question

The enthalpy changes for the following processes are listed below
$	ext{Cl}_2(	ext{g}) = 2	ext{Cl}(	ext{g}), 242.3 	ext{ kJ mol}^{-1}$
$	ext{I}_2(	ext{g}) = 2	ext{I}(	ext{g}), 151.0 	ext{ kJ mol}^{-1}$
$	ext{ICl}(	ext{g}) = 	ext{I}(	ext{g}) + 	ext{Cl}(	ext{g}), 211.3 	ext{ kJ mol}^{-1}$
$	ext{I}_2(	ext{s}) = 	ext{I}_2(	ext{g}), 62.76 	ext{ kJ mol}^{-1}$
Given that the standard states for iodine and chlorine are $	ext{I}_2(	ext{s})$ and $	ext{Cl}_2(	ext{g})$, the standard enthalpy of formation of ICl is

Options:

(A) $-14.6 	ext{ kJ mol}^{-1}$

(B) $-16.8 	ext{ kJ mol}^{-1}$

(C) $+16.8 	ext{ kJ mol}^{-1}$

(D) $+244.8 	ext{ kJ mol}^{-1}$

Sigma Percentile · Accepted Answer

Imagine you are an architect of molecules. Your task is to build exactly one mole of iodine monochloride ($ICl$) gas from its most stable, naturally occurring building blocks. In the world of thermodynamics, this is called the **standard enthalpy of formation** ($\Delta_f H^\circ$).

To do this, we must start with the elements in their standard states at room temperature. Chlorine is a pale green gas ($Cl_2(g)$), but iodine is a dark, crystalline solid ($I_2(s)$). Our target reaction is:

$$ \frac{1}{2}I_2(s) + \frac{1}{2}Cl_2(g) ightarrow ICl(g) $$

### The Magic of Hess's Law

Now, we might not be able to measure the heat of this exact reaction directly in a calorimeter. But thermodynamics gives us a superpower: **Hess's Law of Constant Heat Summation**. Because enthalpy is a state function, it doesn't matter *how* we get from our reactants to our products. The total energy change will be exactly the same whether we do it in one step or a hundred steps.

So, let's design a hypothetical pathway using the data we have!

### Step 1: Sublimation of Iodine

Our first hurdle is that iodine is a solid. We need it to be a gas so it can react freely. We must sublimate it. The problem tells us that turning one mole of solid iodine into gas requires $62.76 	ext{ kJ}$.

$$ I_2(s) ightarrow I_2(g) \quad \Delta H = 62.76 	ext{ kJ mol}^{-1} $$

However, our balanced equation only calls for half a mole of iodine. Therefore, the energy required for this step is exactly half:

$$ \Delta H_1 = \frac{1}{2} 	imes 62.76 = 31.38 	ext{ kJ} $$

### Step 2: Atomization of Gases

Now we have gaseous $I_2$ and $Cl_2$. But they are still diatomic molecules. To form $ICl$, we need to rip these molecules apart into individual, highly reactive atoms. Breaking bonds always requires an input of energy (it's an endothermic process).

The bond dissociation energies are $151.0 	ext{ kJ mol}^{-1}$ for $I_2$ and $242.3 	ext{ kJ mol}^{-1}$ for $Cl_2$. Again, we only need to break half a mole of each:

$$ \Delta H_2 = \frac{1}{2}(151.0) + \frac{1}{2}(242.3) = 75.5 + 121.15 = 196.65 	ext{ kJ} $$

### Step 3: Forging the ICl Bond

We finally have our raw materials: one mole of iodine atoms and one mole of chlorine atoms. It's time to bring them together!

$$ I(g) + Cl(g) ightarrow ICl(g) $$

When a chemical bond forms, the system becomes more stable, and energy is released to the surroundings. The problem states that breaking the $ICl$ bond requires $211.3 	ext{ kJ}$. Therefore, *forming* it must release exactly that same amount of energy. We represent this release with a negative sign:

$$ \Delta H_3 = -211.3 	ext{ kJ} $$

### The Final Calculation

We have successfully navigated our hypothetical pathway. According to Hess's Law, the total standard enthalpy of formation is simply the sum of the enthalpies of our three steps:

$$ \Delta_f H^\circ = \Delta H_1 + \Delta H_2 + \Delta H_3 $$

$$ \Delta_f H^\circ = 31.38 + 196.65 - 211.3 $$

$$ \Delta_f H^\circ = 228.03 - 211.3 = +16.73 	ext{ kJ mol}^{-1} $$

The positive sign tells us that, overall, this is an **endothermic** process. The system absorbed more energy breaking the initial bonds and sublimating the iodine than it released when forming the new $ICl$ bond. Looking at our options, $+16.8 	ext{ kJ mol}^{-1}$ is the closest match. We've cracked the code!

Select Answer:

Visualized Solution

\text{The Target Reaction}

\text{Hess's Law}

\text{Step 1: Sublimation}

\text{Step 2: Atomization}

\text{Step 3: Bond Formation}

\text{Applying Hess's Law}

\text{Calculation}

\text{Final Result}

The Sigma Insight: Enthalpy and Hess's Law

The Magic of Hess's Law

Step 1

Step 2

Step 3

The Final Calculation

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