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?

Short answer

The net reaction is \(\mathrm{F} \longrightarrow \mathrm{B} + \frac{1}{2}\mathrm{D} - \frac{1}{2}\mathrm{C}\), and the overall heat of the reaction is 47.0 kJ/mol.

Step-by-step solution

Adjust the given equations

For the first equation, we need to reverse it and multiply by 1/6: $$ \begin{aligned} -1/6(3 \mathrm{~D} \longrightarrow 3 \mathrm{~A}+6 \mathrm{~B}) & \Longrightarrow \mathrm{D} \longrightarrow \frac{1}{2}\mathrm{A}+\mathrm{B} & \Delta H_1 &= \frac{1}{6}(403 \mathrm{~kJ/mol}) = 67.2 \mathrm{~kJ/mol} \\ \end{aligned} $$ For the second and third equations, we need to divide them by 2: $$ \begin{aligned} 1/2(\mathrm{E}+2 \mathrm{~F} \longrightarrow \mathrm{A}) & \Longrightarrow \frac{1}{2}\mathrm{E}+\mathrm{F} \longrightarrow \frac{1}{2}\mathrm{A} & \Delta H_2 &= \frac{1}{2}(-105.2 \mathrm{~kJ/mol}) = -52.6 \mathrm{~kJ/mol}\\ 1/2(\mathrm{C} \longrightarrow \mathrm{E} + 3 \mathrm{~D}) & \Longrightarrow \frac{1}{2}\mathrm{C} \longrightarrow \frac{1}{2}\mathrm{E} + \frac{3}{2}\mathrm{D} & \Delta H_3 &= \frac{1}{2}(64.8 \mathrm{~kJ/mol}) = 32.4 \mathrm{~kJ/mol} \end{aligned} $$

Add the adjusted equations and cancel out the common species

We will now add the three adjusted equations: $$ \begin{aligned} & \mathrm{D} \longrightarrow \frac{1}{2}\mathrm{A}+\mathrm{B} \\ + & \frac{1}{2}\mathrm{E}+\mathrm{F} \longrightarrow \frac{1}{2}\mathrm{A} \\ + & \frac{1}{2}\mathrm{C} \longrightarrow \frac{1}{2}\mathrm{E} + \frac{3}{2}\mathrm{D} \\ \cline{1-2} & \frac{1}{2}\mathrm{C} + \frac{1}{2}\mathrm{E} + \mathrm{F} + \mathrm{D} \longrightarrow \frac{1}{2}\mathrm{A}+\mathrm{B} + \frac{1}{2}\mathrm{A} + \frac{3}{2}\mathrm{D} \\ \end{aligned} $$ Now, cancel out the common species: $$ \begin{aligned} & \cancel{\frac{1}{2}\mathrm{E}} + \mathrm{F} + \cancel{\mathrm{D}} \longrightarrow \cancel{\frac{1}{2}\mathrm{A}}+\mathrm{B} + \cancel{\frac{1}{2}\mathrm{A}} + \frac{1}{2}\mathrm{D} \\ \end{aligned} $$ The net reaction is: $$ \begin{aligned} \mathrm{F} \longrightarrow \mathrm{B} + \frac{1}{2}\mathrm{D} - \frac{1}{2}\mathrm{C} \\ \end{aligned} $$

Calculate the overall enthalpy change

Sum up the enthalpy changes of the adjusted equations: $$ \Delta H_{net} = \Delta H_1 + \Delta H_2 + \Delta H_3 = 67.2 \mathrm{~kJ/mol} + (-52.6 \mathrm{~kJ/mol}) + 32.4 \mathrm{~kJ/mol} = 47.0 \mathrm{~kJ/mol} $$ The overall heat of the net reaction is 47.0 kJ/mol.

Key concepts

Enthalpy Change Calculations

When studying chemistry, understanding how to calculate the enthalpy change of a reaction is crucial. Enthalpy changes, denoted as \( \Delta H \), are manifestations of energy changes within chemical systems. Hess's Law provides a neat pathway to compute these changes by using known enthalpy differences of other reactions. It states that the enthalpy change of a chemical reaction is the same, regardless of how many steps the reaction is carried through.
Consider the case when different chemical equations form a given reaction sequence. Each equation has its enthalpy change value. By algebraically manipulating these equations, such as reversing or scaling them, you can derive the target equation. Remember, reversing a reaction negates its \( \Delta H \), and scaling a reaction scales \( \Delta H \) similarly. This flexibility showcases the additive nature of enthalpy - as you can add the enthalpy changes of individual steps to ascertain the overall energy change.
To master enthalpy change calculations, focus on balancing each manipulated equation correctly, maintaining stoichiometry. Always check for potential errors by verifying reactants and products alongside provided \( \Delta H \) values.

Chemical Equations

Chemical equations represent reactions using symbols and formulas to detail reactants converting into products. They can be simple or complex, depending on the number and type of chemicals involved. In the equations from the exercise, molecules like \( \mathrm{A} \), \( \mathrm{B} \), and \( \mathrm{D} \) are involved, each with a specified enthalpy change. A balanced chemical equation must have the same number of atoms and charge balance on both sides.
Balancing and manipulating chemical equations are essential skills. Through this, you derive desired reactions by adding, dividing, or reversing equations, akin to adjusting mathematical equations in algebra. For example, reversing a reaction requires changing the products into reactants and vice versa.
Keep a strategic approach:

• Identify key molecules and their enthalpies.
• Manipulate equations to align with the target reaction while maintaining balance.
• Track your steps carefully to ensure nothing is missed or wrongly altered.

Practicing these skills leads to better precision and understanding of chemical dynamics.

Thermodynamics

Thermodynamics forms the backbone of chemistry, focusing on energy transformations. Enthalpy is a part of this broader domain. The exercises students solve often employ thermodynamic principles, like energy conservation or entropy related to reaction feasibilities. Hess's Law is a fundamental principle in this field, reflecting how energy paths can be rearranged as long as initial and final states are consistent.
Think of thermodynamics as the science of puzzle-solving in chemistry - it allows you to predict and understand energy flow and exchange in reactions. Whether a reaction releases energy (exothermic) or absorbs it (endothermic), enthalpy tells part of the thermodynamic story.

• Use energy diagrams to visualize processes.
• Understand state functions - why the route doesn't matter but the start and end do.
• Apply concepts like entropy and Gibbs free energy for deeper insights.

Approaching chemistry through the lens of thermodynamics enables students to understand the 'why' behind reactions, fortifying both theoretical and practical knowledge base.