Problem 23 - Chemistry The Central Science

Question

For each of the following gas-phase reactions, indicate how the rate of disappearance of each reactant is related to the rate of appearance of each product: \(\begin{array}{l}{\text { (a) } \mathrm{H}_{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)} \\ {\text { (b) } 2 \mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)} \\ {\text { (c) } \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)} \\ {\text { (d) } \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{NH}_{3}(g)}\end{array}\)

Step-by-Step Solution

Answer

For each of the given gas-phase reactions: (a) -1 * d[H2O2]/dt = +1 * d[H2]/dt = +1 * d[O2]/dt (b) -2 * d[N2O]/dt = +2 * d[N2]/dt = +1 * d[O2]/dt (c) -1 * d[N2]/dt = -3 * d[H2]/dt = +2 * d[NH3]/dt (d) -1 * d[C2H5NH2]/dt = +1 * d[C2H4]/dt = +1 * d[NH3]/dt

Step 1: (a) Reaction H2O2(g) → H2(g) + O2(g)

In this reaction, we have: 1 mole of H2O2(g) → 1 mole of H2(g) + 1 mole of O2(g). The rate of disappearance of H2O2 is equal to the rate of appearance of H2 and O2, since they are in a 1-to-1 ratio. Therefore, we can write: Rate of disappearance of H2O2 = -d[H2O2]/dt Rate of appearance of H2 = +d[H2]/dt Rate of appearance of O2 = +d[O2]/dt Since they are in a 1-to-1 ratio: -1 * d[H2O2]/dt = +1 * d[H2]/dt = +1 * d[O2]/dt

Step 2: (b) Reaction 2 N2O(g) → 2 N2(g) + O2(g)

In this reaction, we have: 2 moles of N2O(g) → 2 moles of N2(g) + 1 mole of O2(g). The stoichiometric coefficients indicate the relationship between the disappearance of N2O and the appearance of N2 and O2. We can write: Rate of disappearance of N2O = -d[N2O]/dt Rate of appearance of N2 = +d[N2]/dt Rate of appearance of O2 = +d[O2]/dt Since they are in a 2-to-2-to-1 ratio: -2 * d[N2O]/dt = +2 * d[N2]/dt = +1 * d[O2]/dt

Step 3: (c) Reaction N2(g) + 3 H2(g) → 2 NH3(g)

In this reaction, we have: 1 mole of N2(g) + 3 moles of H2(g) → 2 moles of NH3(g). The stoichiometric coefficients indicate the relationship between the disappearance of N2 and H2, and the appearance of NH3. We can write: Rate of disappearance of N2 = -d[N2]/dt Rate of disappearance of H2 = -d[H2]/dt Rate of appearance of NH3 = +d[NH3]/dt Since they are in a 1-to-3-to-2 ratio: -1 * d[N2]/dt = -3 * d[H2]/dt = +2 * d[NH3]/dt

Step 4: (d) Reaction C2H5NH2(g) → C2H4(g) + NH3(g)

In this reaction, we have: 1 mole of C2H5NH2(g) → 1 mole of C2H4(g) + 1 mole of NH3(g). The rate of disappearance of C2H5NH2 is equal to the rate of appearance of C2H4 and NH3, since they are also in a 1-to-1 ratio. We can write: Rate of disappearance of C2H5NH2 = -d[C2H5NH2]/dt Rate of appearance of C2H4 = +d[C2H4]/dt Rate of appearance of NH3 = +d[NH3]/dt Since they are in a 1-to-1-to-1 ratio: -1 * d[C2H5NH2]/dt = +1 * d[C2H4]/dt = +1 * d[NH3]/dt

Key Concepts

Stoichiometry

Stoichiometry is like a recipe for chemical reactions, helping us understand the proportions of reactants and products involved. When you have a balanced chemical equation, it shows the precise amount of each substance needed or produced.

• For instance, in the reaction \( ext{H}_2 ext{O}_2(g) \rightarrow ext{H}_2(g) + ext{O}_2(g)\), the stoichiometric coefficients (the numbers in front of the chemical formulas) are all 1. This means for every 1 mole of \( ext{H}_2 ext{O}_2\) disappearing, we produce 1 mole of \( ext{H}_2\) and 1 mole of \( ext{O}_2\).
• In Equation (b), \(2 ext{N}_2 ext{O}(g) \rightarrow 2 ext{N}_2(g) + ext{O}_2(g)\), the coefficients are 2, 2, and 1. Thus, for every 2 moles of \( ext{N}_2 ext{O}\) consumed, 2 moles of \( ext{N}_2\) and 1 mole of \( ext{O}_2\) are produced. This highlights how stoichiometry governs the proportions.
• Understanding stoichiometry is crucial for predicting how much of each reactant you'll need or how much product you'll get, much like knowing the ingredients required to bake a certain number of cookies.

Gas-Phase Reactions

Gas-phase reactions are those in which all reactants and products are in the gaseous state. They are unique because gases easily expand and interact, allowing reactions to occur over a large volume.

• An advantage of gas-phase reactions is that they can be easily studied in terms of their pressure and temperature, which influence the speed of the reaction.
• For example, when looking at \( ext{C}_2 ext{H}_5 ext{NH}_2(g) \rightarrow ext{C}_2 ext{H}_4(g) + ext{NH}_3(g)\), all species involved are gases. This uniform state can simplify the calculation and analysis of reaction rates.
• In industries, understanding these reactions is vital, like in the Haber process, where nitrogen and hydrogen gases react to form ammonia. The ability of gases to mix can make reactions more efficient.

Rate of Disappearance and Appearance

The rate of disappearance and appearance is about how fast reactants are used up or how fast products form during a reaction.

• This rate is often measured as a change in concentration over time, such as \(-d[ ext{A}]/dt\) for disappearance or \(+d[ ext{B}]/dt\) for appearance.
• Using our reaction \(2 ext{N}_2 ext{O}(g) \rightarrow 2 ext{N}_2(g) + ext{O}_2(g)\), the rate of disappearance of \( ext{N}_2 ext{O}\) is twice that of \( ext{O}_2\) appearance because of the stoichiometric coefficients.
• Getting these rates right is crucial in predicting how long a reaction will take, optimizing conditions, and scaling up processes from a lab to industrial-size reactions.