Question

What will be the enthalpy of combustion of carbon to produce carbon monoxide on the basis of data given below: $$ \begin{aligned} &\mathrm{C}_{(s)}+\mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)}-393.4 \mathrm{~kJ} \\ &\mathrm{CO}_{(g)}+\frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)}-283.0 \mathrm{~kJ} \end{aligned} $$ (a) \(+676.4 \mathrm{~kJ}\) (b) \(-676.4 \mathrm{~kJ}\) (c) \(-110.4 \mathrm{~kJ}\) (d) \(+110.4 \mathrm{~kJ}\)

Short answer

-110.4 kJ

Step-by-step solution

Understand the given reactions

First, we note the two given reactions and their corresponding enthalpies of combustion. The reaction for combustion of carbon to carbon dioxide is: \[ \mathrm{C}_{(s)} + \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)} -393.4 \mathrm{~kJ} \] The reaction for combustion of carbon monoxide to carbon dioxide is: \[ \mathrm{CO}_{(g)} + \frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)} -283.0 \mathrm{~kJ} \]

Write the target reaction

Our goal is to find the enthalpy for the combustion of carbon to produce carbon monoxide. We express this reaction as: \[ \mathrm{C}_{(s)} + \frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{(g)} \]

Apply Hess's Law

According to Hess's Law, the total enthalpy change for a chemical reaction is the same no matter how the reaction occurs, as long as the initial and final conditions are the same. We reverse the second reaction and subtract it from the first to obtain the target reaction, inverting its enthalpy sign in the process: \[ \begin{aligned} &\mathrm{C}_{(s)} + \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)} &-393.4 \mathrm{~kJ} \ -&\mathrm{CO}_{2(g)} \rightarrow \mathrm{CO}_{(g)} + \frac{1}{2} \mathrm{O}_{2(g)} &+283.0 \mathrm{~kJ} \ \hline &\mathrm{C}_{(s)} + \frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{(g)} &-110.4 \mathrm{~kJ} \end{aligned} \]

Infer the enthalpy of combustion

The result of the subtraction gives us the enthalpy change for the target reaction. Hence the enthalpy of combustion of carbon to produce carbon monoxide is \( -110.4 \mathrm{~kJ} \).

Key concepts

Hess's Law

Understanding Hess's Law is crucial for mastering thermochemistry, and it underpins the process of calculating the enthalpy change of a reaction, even when it's not possible to measure it directly. Hess's Law states that the total change in enthalpy for a given chemical reaction is the same, regardless of the number of steps it takes to achieve the reaction, provided the initial and final conditions are consistent.

For students tackling combustion reactions or any other type, the power of this law is in its ability to compute enthalpy changes for reactions that are difficult to perform in a lab setting. By using Hess's Law, we can 'construct' a pathway of known reactions that add up to the overall process we're interested in. For instance, if you want to know the enthalpy change when solid carbon burns to form carbon monoxide, you can use known values for carbon burning to carbon dioxide and carbon monoxide burning to carbon dioxide to find it. The key to Hess's Law is manipulating these equations so they add up to your desired equation, commonly involving reversing reactions and multiplying by coefficients.

Thermochemical Equations

Thermochemical equations are not just about stating what reacts with what—these reactions go a step further to quantify the energy changes involved. Each thermochemical equation is a balanced chemical equation that also includes the enthalpy change, denoted as \( \Delta H \), for the reaction. This can be expressed in kilojoules (kJ) and is usually presented alongside the equation, indicating whether energy is absorbed (+) or released (-) during the reaction.

When dealing with thermochemical equations, it's important to pay attention to the states of the reactants and products as well as the stoichiometry; the coefficients in the balanced equation will tell you how much heat is involved per mole of substance. By understanding the stoichiometry and the concept of enthalpy, you can predict whether a reaction will be exothermic (releasing heat) or endothermic (absorbing heat), which is essential for any chemist.

Standard Enthalpy Change

When we discuss the enthalpy change (\( \Delta H \) of a reaction, we refer to the amount of energy absorbed or released under standard conditions, typically 1 atm and 25°C (298 K). The 'standard' part means that all reactants and products are in their standard states. For gases, the standard state is pure gas at 1 atm; for solids and liquids, it's the pure substance in the most stable form at 1 atm and the specified temperature; and for substances in solution, it is a concentration of 1 M.

The notation \( \Delta H° \) indicates a standard enthalpy change. This means we have a reference point for measuring and comparing the energy changes of reactions. For students, understanding standard enthalpy changes is vital for predicting the feasibility of reactions and for quantifying the energy involved, whether it is being used to produce heat and power in industrial processes or in everyday occurrences like burning fuels or metabolic reactions in biology.