Question

From the enthalpies of reaction $$ \begin{aligned} 2 \mathrm{C}(s)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{CO}(g) & \Delta H=-221.0 \mathrm{~kJ} \\ 2 \mathrm{C}(s)+\mathrm{O}_{2}(g)+4 \mathrm{H}_{2}(g) & \longrightarrow & 2 \mathrm{CH}_{3} \mathrm{OH}(g) & \Delta H=-402.4 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta H\) for the reaction $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g) $$

Short answer

The enthalpy change for the reaction \(CO(g) + 2 H_{2}(g) \longrightarrow CH_{3}OH(g)\) is \(\Delta H = -90.7\ \mathrm{kJ}\).

Step-by-step solution

Identify the desired reaction

Our desired reaction is: $$ \mathrm{CO}(g) + 2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g) $$ We need to manipulate the given reactions to arrive at this final desired reaction.

Manipulate the given reactions to form the desired reaction

First, we notice that the first given reaction has 2 CO(g) being formed on the right-hand side, while in the desired reaction we have only 1 CO(g). So, we will divide the first given reaction by 2: $$ \mathrm{C}(s) + \frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}(g) \qquad \Delta H_{1} = -110.5\ \mathrm{kJ} $$ Now, the second given reaction has 2 CH3OH(g) in the products, but we only need 1 CH3OH(g) for the desired reaction. Divide the second given reaction by 2: $$ \mathrm{C}(s) + \frac{1}{2} \mathrm{O}_{2}(g) + 2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g) \qquad \Delta H_{2} = -201.2\ \mathrm{kJ} $$ At this point, we can see that the sum of these two modified reactions gives us the desired reaction. Accordingly, we will subtract the first modified reaction from the second modified reaction to obtain the desired reaction.

Use Hess's Law to determine the enthalpy of the desired reaction

Now we will apply Hess's Law: $$ (\mathrm{C}(s) + \frac{1}{2} \mathrm{O}_{2}(g) + 2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g)) - (\mathrm{C}(s) + \frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}(g)) $$ The Carbon and Oxygen in the reactants cancel out, giving us the desired reaction: $$ \mathrm{CO}(g) + 2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g) $$ The enthalpy change of the desired reaction is the difference between the enthalpies of the two modified reactions: $$ \Delta H_{\text{desired}} = \Delta H_{2} - \Delta H_{1} = (-201.2\ \mathrm{kJ}) - (-110.5\ \mathrm{kJ}) = -90.7\ \mathrm{kJ} $$

Final answer

The enthalpy change for the desired reaction is -90.7 kJ: $$ \mathrm{CO}(g) + 2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g); \qquad \Delta H = -90.7\ \mathrm{kJ} $$

Key concepts

Enthalpy of Reaction

Enthalpy of reaction refers to the heat change that occurs during a chemical reaction at constant pressure. This change in heat is represented as \( \Delta H \). When a reaction releases heat, it is called exothermic and \( \Delta H \) is negative; conversely, when it absorbs heat, it is endothermic and \( \Delta H \) is positive.
To calculate the enthalpy of a particular reaction from known reactions, we use a principle called Hess's Law. This law writes that if a process can be represented by multiple steps, the total enthalpy change is the sum of the enthalpy changes of the individual steps. So, by manipulating known reactions, we can determine the enthalpy of a unknown reaction.
In our exercise, you notice that the given reactions are rearranged to form the desired equation \( \mathrm{CO}(g) + 2 \mathrm{H}_2(g) \rightarrow \mathrm{CH}_3\mathrm{OH}(g) \). After rearranging, the steps involve subtracting or adding the potential enthalpy changes, resulting in a calculated \( \Delta H \) for the desired reaction.

Chemical Thermodynamics

Chemical thermodynamics is the study of the relationships between heat and chemical reactions. It explores how energy changes in physical processes, especially chemical transformations. A core component of thermodynamics in chemistry involves understanding how reactions evolve under the laws of thermodynamics. Hess's Law, which is applied in this exercise, is fundamental within this field.
There are essential concepts such as energy conservation, spontaneity of reactions, and the balancing of energy throughout different reaction pathways. In the context of our exercise, the focus is on using known enthalpy changes of reactions to derive the enthalpy change of an unknown reaction pathway. Understanding the energy changes helps predict whether a reaction will occur spontaneously or if it requires input energy.
For learners, grasping how thermodynamics connect to enthalpy within chemical reactions is crucial to solving problems that involve predicting energy flow and conservation in chemical processes.

Reaction Manipulation

Reaction manipulation involves rearranging, reversing, or scaling reactions to develop a desired reaction or pathway. This is particularly essential in applying Hess's Law. By adjusting coefficients or altering the direction of reactions, we ensure that we align all chemical species so that unwanted reactants or products cancel out. Each manipulation impacts the enthalpy directly, altering it by a multiplication, division, or sign change accordingly.
In our exercise, manipulation was key to deriving the desired reaction from the provided ones. We noticed we had to divide the coefficients of the given reactions by two to match the mole ratios in our target reaction. This careful adjustment is necessary to ensure the stoichiometry of the reaction is correct before combining the enthalpies of the component reactions.
By mastering reaction manipulation, students can efficiently utilize known reactions to create balanced equations that reflect the processes of unknown reactions, providing a pathway to deduce their enthalpy changes accurately. Understanding this concept allows for strategic adjustments in arriving at a specific chemical equation.