Problem 74 - Chemistry The Molecular Nature of Matter

Question

One thermochemical equation for the reaction of carbon monoxide with oxygen is $$ 3 \mathrm{CO}(g)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{CO}_{2}(g) \quad \Delta H^{\circ}=-849 \mathrm{~kJ} $$ (a) Write the thermochemical equation for the reaction of \(2.00 \mathrm{~mol}\) of \(\mathrm{CO}\) (b) What is the \(\Delta H^{\circ}\) for the reaction that produces \(1.00 \mathrm{~mol}\) of \(\mathrm{CO}_{2} ?\)

Step-by-Step Solution

Answer

The thermochemical equation for the reaction involving 2.00 mol of CO is \(2 \mathrm{CO}(g) + \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g), \Delta H^\circ = -566 \mathrm{~kJ}\). The \(\Delta H^\circ\) for the reaction that produces 1.00 mol of \(\mathrm{CO}_{2}\) is -283 kJ.

Step 1: Write the thermochemical equation for 2 mol CO

The original equation shows the reaction and enthalpy change for 3 moles of CO. To find the thermochemical equation for 2 moles of CO, reduce all coefficients in the original reaction by a factor of 3/2.

Step 2: Adjust the equation's coefficients

Divide the number of moles of each reactant and product by 3/2 to get the new coefficients: 3 CO / (3/2) = 2 CO, (3/2) O2 / (3/2) = O2, and 3 CO2 / (3/2) = 2 CO2.

Step 3: Calculate the enthalpy change for the new equation

Since the enthalpy change \(\Delta H^\circ = -849 \, \mathrm{kJ}\) is for 3 moles of CO, for 2 moles of CO, we divide it by 3/2 (the same factor by which we adjusted the coefficients): \(\Delta H^\circ = -849 \, \mathrm{kJ} \/ (3/2)\).

Step 4: Scale the enthalpy change accordingly

Perform the actual division to find the enthalpy change for 2 moles of CO: \(\Delta H^\circ = -849 \, \mathrm{kJ} \/ 1.5 = -566 \, \mathrm{kJ}\).

Step 5: Write the thermochemical equation for 1 mol CO2

Now to find the enthalpy change for the production of 1 mole of CO2, we scale the enthalpy for the production of 2 moles of CO2 down to 1 mole. Since 2 moles required -566 kJ, 1 mole will require half of that.

Step 6: Calculate the enthalpy change for producing 1 mol CO2

Divide the enthalpy change of 2 moles by 2 to find the enthalpy change for 1 mole of CO2: \(\Delta H^\circ = -566 \, \mathrm{kJ} / 2 = -283 \, \mathrm{kJ}\).

Key Concepts

Enthalpy Change

Enthalpy change, represented by the symbol \( \Delta H \), is the amount of heat absorbed or released during a chemical reaction at constant pressure. This value is critical in thermochemistry, as it helps scientists and engineers understand how much energy is involved in a reaction. In our exercise, the enthalpy change is negative, \( \Delta H^\circ = -849 \mathrm{kJ} \), indicating an exothermic reaction where heat is released into the surroundings.

Understanding how to adjust this value for different amounts of reactants is vital. For instance, the original equation gives us the enthalpy change for 3 moles of carbon monoxide (CO). To find the change for 2 moles, we scale the \( \Delta H \) by the same factor as the change in moles, resulting in a new enthalpy change for the adjusted amount of CO.

Stoichiometry

Stoichiometry involves the quantitative relationships between the reactants and products in a chemical reaction. It's all about the ratios and coefficients in balanced chemical equations. When we modify a thermochemical equation to find out the enthalpy change for a different number of moles, we're applying stoichiometry.

To adjust the coefficients proportionately, divide them by the same factor that relates the moles of CO in the exercise from the initial to the desired amount. Here, we took the coefficients and divided by 3/2 to convert the initial equation for 3 moles of CO to an equation for 2 moles. By mastering the stoichiometric calculations, we ensure that chemical equations are balanced in both mass and energy, crucial for predicting reaction yields and energy changes in various contexts.

Chemical Reactions

Chemical reactions involve rearrangements of atoms as reactants convert into products. Writing and balancing chemical equations are fundamental skills in chemistry, as they represent the conservation of mass and the stoichiometry of reactions. The given exercise involves a combustion reaction, where carbon monoxide reacts with oxygen to produce carbon dioxide.

In such reactions, stoichiometry allows us to calculate not only the amount of each substance required or produced but also the energy change associated with the reaction. By understanding the nature of a reaction—whether it's exothermic like our example, or endothermic, where energy is absorbed—students can predict the energy requirements or releases in industrial processes, environmental systems, and more.