Problem 105 - Chemistry An Atoms First Approach

Question

Consider the following equations: $$\begin{aligned}3 \mathrm{A}+6 \mathrm{B} \longrightarrow 3 \mathrm{D} & & \Delta H=-403 \mathrm{kJ} / \mathrm{mol} \\\\\mathrm{E}+2 \mathrm{F} \longrightarrow \mathrm{A} & & \Delta H=-105.2 \mathrm{kJ} / \mathrm{mol} \\\\\mathrm{C} \longrightarrow \mathrm{E}+3 \mathrm{D} & & \Delta H=64.8 \mathrm{kJ} / \mathrm{mol}\end{aligned}$$ Suppose the first equation is reversed and multiplied by \(\frac{1}{6},\) the second and third equations are divided by \(2,\) and the three adjusted equations are added. What is the net reaction and what is the overall heat of this reaction?

Step-by-Step Solution

Answer

The net reaction is: \(\frac{1}{2}\mathrm{E}+\mathrm{F}+\frac{1}{2}\mathrm{C} \longrightarrow \mathrm{A}+2\mathrm{B}+\frac{3}{2}\mathrm{D}\) and the overall heat of this reaction is \(\Delta H_\text{net} = 47 \mathrm{kJmol}^{-1}\).

Step 1: Reverse first equation and multiply by 1/6

Reversing first equation: $$3 \mathrm{D} \longrightarrow 3 \mathrm{A} + 6 \mathrm{B}$$ Now, multiply by 1/6: \(\frac{1}{6} (3 \mathrm{D} \longrightarrow 3 \mathrm{A} + 6 \mathrm{B})\) $$\mathrm{D} \longrightarrow \mathrm{A}+2 \mathrm{B}$$ After reversing, the enthalpy change will also change its sign. And then we need to multiply it by 1/6 as well: \(\Delta H = (1/6) \times 403 \mathrm{kJmol}^{-1} = 67.2 \mathrm{kJmol}^{-1}\)

Step 2: Divide second and third equations by 2

Divide second equation by 2: \(\frac{1}{2} (\mathrm{E} + 2 \mathrm{F} \longrightarrow \mathrm{A})\) $$\frac{1}{2} \mathrm{E} + \mathrm{F} \longrightarrow \frac{1}{2} \mathrm{A}$$ Divide enthalpy change for this equation by 2: \(\Delta H = -105.2 \mathrm{kJ} / \mathrm{mol} / 2 = -52.6 \mathrm{kJ} / \mathrm{mol}\) Divide third equation by 2: \(\frac{1}{2} (\mathrm{C} \longrightarrow \mathrm{E} + 3 \mathrm{D})\) $$\frac{1}{2} \mathrm{C} \longrightarrow \frac{1}{2} \mathrm{E} + \frac{3}{2} \mathrm{D}$$ Divide enthalpy change for this equation by 2: \(\Delta H = 64.8 \mathrm{kJ} / \mathrm{mol} / 2 = 32.4 \mathrm{kJ} / \mathrm{mol}\)

Step 3: Add the adjusted equations

Now, we need to add the adjusted equations to obtain the net reaction: $$\mathrm{D} \longrightarrow \mathrm{A}+2 \mathrm{B}$$ $$\frac{1}{2} \mathrm{E} + \mathrm{F} \longrightarrow \frac{1}{2} \mathrm{A}$$ $$\frac{1}{2} \mathrm{C} \longrightarrow \frac{1}{2} \mathrm{E} + \frac{3}{2} \mathrm{D}$$ Adding these equations, we get: \[\frac{1}{2}\mathrm{E}+\mathrm{F}+\frac{1}{2}\mathrm{C} \longrightarrow \frac{1}{2}\mathrm{A}+2\mathrm{B}+\frac{1}{2}\mathrm{A}+\frac{3}{2}\mathrm{D}\] Simplifying the equation: \[\frac{1}{2}\mathrm{E}+\mathrm{F}+\frac{1}{2}\mathrm{C} \longrightarrow \mathrm{A}+2\mathrm{B}+\frac{3}{2}\mathrm{D}\] Now, let's find the overall heat by adding the enthalpy changes: $$\Delta H_\text{net} = 67.2 \mathrm{kJmol}^{-1} - 52.6 \mathrm{kJmol}^{-1} + 32.4 \mathrm{kJmol}^{-1} = 47 kJ/mol$$

Step 4: Final Answer

The net reaction is: \[\frac{1}{2}\mathrm{E}+\mathrm{F}+\frac{1}{2}\mathrm{C} \longrightarrow \mathrm{A}+2\mathrm{B}+\frac{3}{2}\mathrm{D}\] The overall heat of this reaction is: \(\Delta H_\text{net} = 47 \mathrm{kJmol}^{-1}\)

Key Concepts

Chemical Reaction Equations

A chemical reaction equation represents the transformation of reactants into products, showcasing the conservation of mass by balancing atoms on both sides of the equation. In straightforward terms, this equation tells us what reacts (reactants) and what is formed (products) during a chemical reaction.

For instance, in the given exercise, we start with the equation:
\[3 \mathrm{A} + 6 \mathrm{B} \longrightarrow 3 \mathrm{D}\]
This implies that three molecules of substance A react with six molecules of substance B to yield three molecules of substance D. The \(\Delta H\) value associated with this reaction informs us about the heat change, which in this case is -403 kJ/mol, indicating that the reaction is exothermic, releasing heat into the surroundings.

Understanding chemical reaction equations is essential because they not only provide the stoichiometry —the quantitative relationship between reactants and products in a chemical reaction— but also because they can show us the energy changes that occur, which is a critical aspect in understanding thermodynamics in chemistry.

Reversing Chemical Equations

In thermodynamics, the direction of a chemical reaction can be reversed, which also flips the sign of the enthalpy change \(\Delta H\). This is essential for the concept of Hess's Law, which we will discuss in the next section.

Reversing the chemical equation:
\[3 \mathrm{D} \longrightarrow 3 \mathrm{A} + 6 \mathrm{B}\]
If we think of a chemical reaction as a journey from point A to point B (reactants to products), reversing the reaction is like going back from point B to point A. As part of this reversal, if the original process released energy, the reverse will require that same amount of energy to proceed. So, if the original reaction in the exercise had a \(\Delta H\) of -403 kJ/mol, reversing it would change the enthalpy to +403 kJ/mol. Applying a stoichiometric factor, such as \(\frac{1}{6}\) in our exercise, the enthalpy change is scaled accordingly to reflect the adjusted amount of substance involved in the reaction.

It's like dividing a cake into smaller pieces; the total amount of cake (energy) remains the same, but the size of each piece (enthalpy change per mole) gets smaller. This concept is pivotal when performing calculations involving enthalpy changes in various chemical scenarios.

Hess's Law

Hess's Law is a cornerstone of chemical thermodynamics, stating that the total enthalpy change for a reaction is the same, no matter how many steps or stages the reaction is carried out in. This law is based on the fact that enthalpy is a state function, meaning that its value depends only on the initial and final states of the system, not on the path taken to get from one to the other.

In the context of the exercise:
\[\Delta H_\text{net} = 67.2 \mathrm{kJ/mol} - 52.6 \mathrm{kJ/mol} + 32.4 \mathrm{kJ/mol} = 47 \mathrm{kJ/mol}\]
By adjusting and adding up individual reactions and their respective enthalpy changes, we can calculate the overall enthalpy change for the net reaction. This application of Hess's Law not only simplifies complex reactions into simpler steps but also allows us to use known enthalpies of formation or reaction to determine unknown values.

Remember, the central idea is that the total heat change (the energy transferred as heat during the chemical reaction at a constant pressure) is constant for a reaction, regardless of the route by which the chemical transformation occurs. This principle is incredibly useful in chemistry, especially in cases where a reaction is too difficult to perform or measure directly.

Hess's Law is a powerful tool in the hands of chemists, allowing us to harness the laws of thermodynamics and stoichiometry to predict energy changes and thus understand and predict chemical behavior.