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When mixed at \(700^{\circ} \mathrm{K}, \mathrm{H}_{2}(\mathrm{~g})\) and \(\mathrm{I}_{2}\) (g) react to produce \(\mathrm{HI}(\mathrm{g})\). The reaction is first order in \(\mathrm{H}_{2}\) and first order in \(\mathrm{I}_{2}\). Suppose at time \(t=0\), one mole of \(\mathrm{H}_{2}\) and one mole of \(\mathrm{I}_{2}\) are simultaneously injected into a 1 -liter box. One second later, before the reaction is complete, the contents of the box are examined for the number of moles of HI. What would be the probable effect on this number if each of the following changes were made in the initial conditions? (a) Use two moles of \(\mathrm{H}_{2}\) instead of one. (b) Use two moles of \(\mathrm{I}_{2}\) instead of one. (c) Use a 2-liter box. (d) Raise the temperature to \(750^{\circ} \mathrm{K}\). (e) Add a platinum catalyst, (f) Add enough neon gas to double the initial pressure.

Short Answer

Expert verified
(a) Using two moles of H₂ will result in a higher concentration of HI after 1 second. (b) Using two moles of I₂ will also result in a higher concentration of HI after 1 second. (c) Using a 2-liter box will result in a lower concentration of HI after 1 second. (d) Raising the temperature to \(750^{\circ} \mathrm{K}\) will result in a higher concentration of HI after 1 second. (e) Adding a platinum catalyst will result in a higher concentration of HI after 1 second. (f) Adding enough neon gas to double the initial pressure may result in a slight increase in the concentration of HI after 1 second.

Step by step solution

01

(a) Use two moles of \(\mathrm{H}_{2}\) instead of one

Doubling the concentration of H₂ will increase the rate of the reaction since it is first order with respect to H₂. As a result, the concentration of HI will be higher after 1 second compared to the initial conditions.
02

(b) Use two moles of \(\mathrm{I}_{2}\) instead of one

Similarly, doubling the concentration of I₂ will also increase the rate of the reaction as it is first order with respect to I₂. This will lead to a higher concentration of HI after 1 second compared to the initial conditions.
03

(c) Use a 2-liter box

Using a 2-liter box will decrease the concentrations of H₂ and I₂, as the volume has doubled. Since the reaction is first order with respect to both reactants, the rate will decrease, resulting in a lower concentration of HI after 1 second compared to the initial conditions.
04

(d) Raise the temperature to \(750^{\circ} \mathrm{K}\)

Raising the temperature generally increases the reaction rate, as molecules have more energy to overcome the activation energy barrier. This will lead to a higher concentration of HI after 1 second compared to the initial conditions.
05

(e) Add a platinum catalyst

Adding a catalyst lowers the activation energy of the reaction, increasing the rate. This will result in a higher concentration of HI after 1 second compared to the initial conditions.
06

(f) Add enough neon gas to double the initial pressure

Adding neon gas will increase the pressure but will not directly affect the chemical reaction since neon is an inert gas. However, the increased pressure may have an indirect effect on the rates of the reactants. With the gaseous species both first order, higher pressure improves the probability of collisions between H₂ and I₂. Thus, there may be a slight increase in the concentration of HI after 1 second compared to the initial conditions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reaction Rates
Understanding reaction rates is crucial when exploring how fast a chemical reaction occurs. It refers to the change in the concentration of reactants or products per unit time. A faster reaction rate means that the reactants are being consumed or the products are being formed more quickly. The rate can be affected by various factors, such as concentration, temperature, and the presence of a catalyst.

For instance, in the exercise provided, when the concentration of either \(\text{H}_2\) or \(\text{I}_2\) is doubled, the reaction rate increases, producing more \(\text{HI}\) in one second. Reaction rates are a foundational concept in chemical kinetics, as they help to predict how changes in conditions can affect the progress of a chemical reaction.
Order of Reaction
The order of reaction is a term used to express the dependency of the reaction rate on the concentration of reactants. It is given as an exponent in the rate equation. A reaction being ‘first order’ in a reactant means that the rate of the reaction is directly proportional to the concentration of that one reactant, while other conditions remain constant.

In the exercise, the reaction is first order with respect to both \(\text{H}_2\) and \(\text{I}_2\). This implies that changing the amount of either reactant directly affects the rate of production of \(\text{HI}\). If the concentration of \(\text{H}_2\) or \(\text{I}_2\) is doubled, the rate of the reaction is expected to double as well.
Catalysts
Catalysts are substances that increase the rate of a chemical reaction without being consumed in the process. They achieve this by providing a different pathway for the reaction with a lower activation energy. This alternative pathway allows more particles to have enough energy to react when they collide, thus increasing the reaction rate.

In the given problem, adding a platinum catalyst to the reaction mixture increases the production of \(\text{HI}\) after one second. The catalyst does not affect the overall stoichiometry or the equilibrium of the reaction; it simply helps the reactants to convert into products more efficiently.
Chemical Equilibrium
Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of the reactants and products remain constant over time. It’s crucial to note that equilibrium does not mean that the reactants and products are present in equal amounts, but rather that their ratios no longer change.

While the exercise provided does not directly explore equilibrium, understanding this concept is essential. A change in temperature, pressure, or concentration can disturb the equilibrium, and the system will adjust according to Le Chatelier's principle to re-establish equilibrium. This concept is closely related to how reaction conditions are manipulated to favor the formation of desired products.
Reaction Mechanisms
A reaction mechanism is a detailed step-by-step description of how a chemical reaction proceeds at the molecular level. It involves a series of elementary steps that include bond breaking and forming, which lead to the overall reaction. The speed of a reaction is determined by the slowest step, also known as the rate-determining step.

Identifying the mechanism can explain why certain factors affect the reaction rate. For example, in our exercise, temperature and catalysts are external factors that influence the rate, while the order of reaction provides insight into the likely mechanism of the process. By studying reaction mechanisms, scientists can better understand and control chemical processes.

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Most popular questions from this chapter

When heated to \(600^{\circ} \mathrm{C}\), acetone \(\left(\mathrm{CH}_{3} \mathrm{COCH}_{3}\right)\) decomposes to give . CO and various hydrocarbons. The reaction is found to be first order in acetone concentration with a half- life of 81 sec. Given at \(600^{\circ} \mathrm{C}\) a 1-liter container into which acetone is injected at \(0.48 \mathrm{~atm}\), approximate how long would it take for the acetone pressure to drop to \(0.45 \mathrm{~atm}\) ?

lwo ditterent molecules, \(\mathrm{A}\) and \(\mathrm{B}\), were reacted to give products according to the equation \(\mathrm{A}+\mathrm{B} \rightarrow\) products The rate of reaction was measured for fixed concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) and the following data was obtained:$$ \begin{aligned} &\begin{array}{|l|l|l|l|} \hline \text { Experiment } & \begin{array}{l} {[\mathrm{A}]} \\ \text { (moles /iter) } \end{array} & \begin{array}{l} \text { [B] } \\ \text { (moles/iter) } \end{array} & \begin{array}{l} \text { Rate } \\ \text { (moles } \left.^{2} \text { /liter }^{2}-\text { sec }\right) \end{array} \\ \hline 1 & 1.0 & 1.0 & 0.05 \\ \hline 2 & 1.0 & 2.0 & 0.10 \\ \hline 3 & 3.0 & 1.0 & 0.45 \\ \hline \end{array}\\\ &\text { Write an expression for the rate of reaction. } \end{aligned} $$

Using a specific case, show that the effect of a \(10^{\circ} \mathrm{K}\) rise in temperature will have a greater effect on the rate constant, \(k\), at low temperatures than it does at high temperatures.

In the reaction \(\mathrm{N}_{2} \mathrm{O}_{5} \rightarrow \mathrm{N}_{2} \mathrm{O}_{4}+(1 / 2) \mathrm{O}_{2}\), the \(\mathrm{N}_{2} \mathrm{O}_{5}\) decomposes by a first-order mechanism. At \(298^{\circ} \mathrm{K}\), the half- life is 340 minutes. Find the value of the reaction rate constant. Calculate the number of minutes required for the reaction to proceed 70 percent towards completion.

\begin{aligned} &\text { Given the consecutive reaction A } \mathrm{k}_{1} \rightarrow \mathrm{B} \mathrm{k}_{2} \rightarrow \mathrm{C} \text { with } \mathrm{k}_{1}=\mathrm{k}_{2} \text { . }\\\ &\text { Draw a graph for the time variation of the concentrations of }\\\ &\mathrm{A}, \mathrm{B}, \text { and } \mathrm{C} \text { . } \end{aligned}

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