Problem 106 - Chemistry The Molecular Nature of Matter

Question

Consider the following thermochemical equations: \(\begin{array}{ll}\mathrm{CH}_{3} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{HCHO}_{2}(l)+\mathrm{H}_{2} \mathrm{O}(l) & \Delta H^{\circ}=-411 \mathrm{~kJ} \\ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(l) & \Delta H^{\circ}=-128 \mathrm{~kJ} \\ \mathrm{HCHO}_{2}(l) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \Delta H^{\circ}=-33 \mathrm{~kJ}\end{array}\) Suppose the first equation is reversed and divided by 2 , The second and third equations are multiplied by \(\frac{1}{2},\) and then the three adjusted equations are added. What is the net reaction, and what is the value of \(\Delta H^{\circ}\) for the net reaction?

Step-by-Step Solution

Answer

\(\Delta H^{\circ}\) for the net reaction \(\mathrm{H}_{2}(g) \longrightarrow \frac{1}{2}\mathrm{O}_{2}(g)\) is \(\frac{411}{2} - \frac{128}{2} - \frac{33}{2} = 125\) kJ.

Step 1: Reverse and halve the first equation

When reversing the first reaction, we invert the products and reactants, and the sign of \(\Delta H^{\circ}\) also changes. So, it becomes: \(\frac{1}{2}\mathrm{CH}_{3}\mathrm{OH}(l) + \frac{1}{2}\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \frac{1}{2}\mathrm{O}_{2}(g) + \frac{1}{2}\mathrm{HCHO}_{2}(l)\), with \(\Delta H^{\circ} = +\frac{411}{2}\mathrm{kJ}\).

Step 2: Halve the second equation

Multiplying the second equation by \(\frac{1}{2}\) gives: \(\frac{1}{2}\mathrm{CO}(g) + \mathrm{H}_{2}(g) \longrightarrow \frac{1}{2}\mathrm{CH}_{3}\mathrm{OH}(l)\), with \(\Delta H^{\circ} = -\frac{128}{2}\mathrm{kJ}\).

Step 3: Halve the third equation

Multiplying the third equation by \(\frac{1}{2}\) gives: \(\frac{1}{2}\mathrm{HCHO}_{2}(l) \longrightarrow \frac{1}{2}\mathrm{CO}(g) + \frac{1}{2}\mathrm{H}_{2}\mathrm{O}(l)\), with \(\Delta H^{\circ} = -\frac{33}{2}\mathrm{kJ}\).

Step 4: Add the adjusted equations

When adding the adjusted equations, some substances will appear on both sides of the equation and thus can be canceled out. After canceling, the net result will be: \(\mathrm{H}_{2}(g) \longrightarrow \frac{1}{2}\mathrm{O}_{2}(g)\).

Step 5: Calculate the net \(\Delta H^{\circ}\) for the reaction

Add up the \(\Delta H^{\circ}\) values from the adjusted equations: \(\Delta H^{\circ} = \frac{411}{2} - \frac{128}{2} - \frac{33}{2}\) kJ. Simplify to find the net \(\Delta H^{\circ}\).

Key Concepts

Enthalpy Change

Enthalpy change is a crucial concept in thermodynamics that relates to the heat exchange in a system during a chemical reaction under constant pressure. It's represented by the symbol \( \Delta H^\circ \).

When a reaction releases heat into its surroundings, it is said to be exothermic, and \( \Delta H^\circ \) will have a negative value. This indicates that the energy of the products is lower than that of the reactants. Conversely, an endothermic reaction absorbs heat, which results in a positive \( \Delta H^\circ \). Here, the products have higher energy compared to the reactants.

Understanding enthalpy change is essential because it helps predict whether a reaction is likely to occur spontaneously. It also conveys how much energy is needed for a reaction to proceed, or how much might be released.

Chemical Reactions

Chemical reactions involve the breaking and forming of chemical bonds between atoms, leading to the transformation of reactants into products. They are described by chemical equations that detail the starting substances, the final substances, and sometimes the states of these substances (such as solid, liquid, gas).

During a chemical reaction, the conservation of mass principles states that atoms are neither created nor destroyed. This balancing act ensures that the same number of atoms of each element is present both before and after the reaction. Chemical reactions can be of various types, such as synthesis, decomposition, single replacement, or double replacement, depending on the nature of the interaction between reactants.

Stoichiometry

Stoichiometry is the quantitative relationship between the reactants and products in a chemical reaction. This discipline is founded on the law of conservation of mass and the concept of the mole, which is a measure of quantity in chemistry.

Using stoichiometry, chemists can calculate how much reactant is needed to produce a certain amount of product, or conversely, how much product can be produced from a given amount of reactant.

Ratio and Proportions

Stoichiometric coefficients in a balanced equation indicate the ratio in which substances react or form products. These coefficients are used to compute the moles of each substance involved.

Limiting Reagents and Yield

Additionally, stoichiometry is critical for identifying the limiting reagent in a reaction – the substance that will be exhausted first, thus determining the amount of product formed. It is also applied when calculating theoretical yield, the maximum quantity of product that could be generated from a given amount of reactant under perfect conditions.