The enthalpies of combustion of carbon and carbon monoxide are -393.5 and -283 kJ mol^-1, respective

Q: Why is the enthalpy of formation of $O_2$ not given in the problem?

By convention, the standard enthalpy of formation for any element in its most stable standard state (like $O_2$ gas, $C$ solid graphite, $H_2$ gas) is exactly zero. It's our baseline reference point.

Q: Can I just subtract the two numbers without writing the equations?

While you might get the right answer by guessing, writing the equations ensures you don't make a sign error or miss a stoichiometric coefficient. It's the only way to guarantee you are applying Hess's Law correctly.

Q: What does the negative sign in the final answer mean?

A negative enthalpy change ($\Delta H < 0$) means the reaction is exothermic. Energy is released into the surroundings when carbon and oxygen combine to form carbon monoxide.

Q: Why do we use a fractional coefficient like $\frac{1}{2} O_2$?

Enthalpy of formation is strictly defined as the energy change when exactly ONE mole of a substance is formed. To balance the equation for forming exactly 1 mole of $CO$, we must use half a mole of $O_2$.

Q: Why does Hess's Law even work?

Hess's Law works because enthalpy is a 'state function'. This means the change in enthalpy depends only on the initial reactants and the final products, not on the specific pathway or number of steps taken to get there.

Question

The enthalpies of combustion of carbon and carbon monoxide are $-393.5$ and $-283 	ext{ kJ mol}^{-1}$, respectively. The enthalpy of formation of carbon monoxide per mole is

Options:

(A) $110.5 	ext{ kJ}$

(B) $676.5 	ext{ kJ}$

(C) $-676.5 	ext{ kJ}$

(D) $-110.5 	ext{ kJ}$

Sigma Percentile · Accepted Answer

### The Magic of Hess's Law

Imagine you are standing at the base of a mountain and you want to reach the peak. You could take a steep, direct trail straight to the top, or you could take a more scenic route that stops at a beautiful lake halfway up before continuing to the summit. Regardless of which path you choose, your total change in altitude from the base to the peak is exactly the same.

This simple idea is the heart of **Hess's Law** in thermodynamics. It states that the total enthalpy change for a chemical reaction is independent of the pathway taken, as long as the initial and final states are the same. Because enthalpy ($H$) is a **state function**, we can treat chemical equations like algebraic equations—adding, subtracting, and multiplying them to find the energy changes of reactions that might be difficult to measure directly.

### Analyzing the Setup

In this problem, we are given two pieces of information:
1. The enthalpy of combustion of carbon: 
   $$C(s) + O_2(g) ightarrow CO_2(g) \quad \Delta H_1 = -393.5 	ext{ kJ mol}^{-1}$$
2. The enthalpy of combustion of carbon monoxide: 
   $$CO(g) + \frac{1}{2}O_2(g) ightarrow CO_2(g) \quad \Delta H_2 = -283.0 	ext{ kJ mol}^{-1}$$

Our goal is to find the enthalpy of formation of carbon monoxide. The standard formation reaction for $CO(g)$ from its elements in their standard states is:
$$C(s) + \frac{1}{2}O_2(g) ightarrow CO(g) \quad \Delta H_f = ?$$

### The Master Equation

Let's visualize this as a two-step journey. The direct path is burning carbon all the way to carbon dioxide ($\Delta H_1$). The two-step path involves first forming carbon monoxide ($\Delta H_f$), and then burning that carbon monoxide to form carbon dioxide ($\Delta H_2$).

According to Hess's Law, the energy of the direct path must equal the sum of the energies of the two-step path:
$$\Delta H_1 = \Delta H_f + \Delta H_2$$

Alternatively, we can manipulate the chemical equations algebraically. If we take the first equation and subtract the second equation from it, we get:
$$[C(s) + O_2(g)] - [CO(g) + \frac{1}{2}O_2(g)] ightarrow CO_2(g) - CO_2(g)$$

Rearranging this by moving the negative terms to the other side yields our exact target equation:
$$C(s) + \frac{1}{2}O_2(g) ightarrow CO(g)$$

### Final Calculation

Since subtracting the equations gave us the target reaction, we simply subtract their enthalpy values:
$$\Delta H_f = \Delta H_1 - \Delta H_2$$

Substituting the given values:
$$\Delta H_f = -393.5 - (-283.0)$$
$$\Delta H_f = -393.5 + 283.0$$
$$\Delta H_f = -110.5 	ext{ kJ mol}^{-1}$$

The negative sign indicates that the formation of carbon monoxide is an **exothermic** process, releasing $110.5 	ext{ kJ}$ of energy per mole. This elegant application of Hess's Law allows us to determine the energy of a reaction without having to perform it in a calorimeter!

Animated Solution for Chemistry - Chemical Thermodynamics: The enthalpies of combustion of carbon and carbon monoxide are $- 393.5$ and $- 283 kJ mol^{- 1}$ , respectively. The enthalpy of formation of carbon monoxide per mole is

Select Answer:

Visualized Solution

\text{Hess's Law Cycle}

\text{Enthalpy of Combustion of Carbon}

\text{Enthalpy of Combustion of CO}

\text{Enthalpy of Formation of CO}

\text{Applying Hess's Law}

\text{Solving for } \Delta H_f^\circ(CO)

\text{Final Result}

\text{Alternative Method: Algebraic Equations}

The Sigma Insight: Enthalpy and Hess's Law

The Magic of Hess's Law

Analyzing the Setup

The Master Equation

Final Calculation

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