Problem 92 - Chemistry The Molecular Science

Question

The statements below relate to this reaction: $$ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HI}(\mathrm{g}) \quad \text { Rate }=k\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right] $$ Determine which statements are true. If a statement is false, indicate why it is incorrect. (a) The reaction must occur in a single step. (b) This is a second-order reaction overall. (c) Raising the temperature will cause the value of \(k\) to decrease. (d) Raising the temperature lowers the activation energy for this reaction. (e) If the concentrations of both reactants are doubled, the rate will double. (f) Adding a catalyst in the reaction will cause the initial rate to increase.

Step-by-Step Solution

Answer

Statements (b) and (f) are true. Statement (a) is inconclusive; (c), (d), and (e) are false.

Step 1: Statement (a) Analysis

Considering the rate law \( \text{Rate}=k[\mathrm{H}_{2}][\mathrm{I}_2] \), it suggests that the reaction relies on the concentrations of \( \mathrm{H}_{2} \) and \( \mathrm{I}_2 \) simultaneously, which is consistent with a single-step reaction mechanism. While the rate law aligns with a single step, it doesn't necessarily prove that it must occur in one step due to complex reaction mechanisms that may also match this rate law.

Step 2: Statement (b) Analysis

To determine the reaction order, the exponents of the reactants in the rate law are added. Here, the reaction follows the rate law \( \text{Rate}=k[\mathrm{H}_{2}][\mathrm{I}_2] \). The order with respect to \( \mathrm{H}_2 \) is 1 and for \( \mathrm{I}_2 \) is 1. Therefore, the overall reaction order is \( 1+1=2 \), confirming that it is a second-order reaction.

Step 3: Statement (c) Analysis

The rate constant \( k \) is related to temperature by the Arrhenius equation \( k = Ae^{-\frac{E_a}{RT}} \), where \( E_a \) is the activation energy. As the temperature increases, \( e^{-\frac{E_a}{RT}} \) also increases, causing \( k \) to increase, not decrease.

Step 4: Statement (d) Analysis

Raising temperature does not lower the activation energy \( E_a \); it affects the rate constant \( k \) as described in the Arrhenius equation. The activation energy remains constant for a given reaction unless a catalyst is added.

Step 5: Statement (e) Analysis

Doubling the concentration of each reactant \( [\mathrm{H}_2] \) and \( [\mathrm{I}_2] \) in the second-order reaction \( \text{Rate}=k[\mathrm{H}_{2}][\mathrm{I}_2] \) results in the rate increasing by a factor of four \( (2 \times 2 = 4) \), not simply doubling. Hence, the statement is false.

Step 6: Statement (f) Analysis

Adding a catalyst provides an alternative pathway with a lower activation energy, increasing the rate of the reaction. Therefore, the initial rate of the reaction will increase with a catalyst.

Key Concepts

Rate Laws

In chemical kinetics, rate laws express the relationship between the rate of a chemical reaction and the concentration of its reactants. For the reaction given, \[\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \longrightarrow 2\mathrm{HI}(\mathrm{g}) \]the rate law is specified as:\[\text{Rate} = k[\mathrm{H}_{2}][\mathrm{I}_{2}]\]This indicates that the reaction rate is directly proportional to the concentrations of hydrogen gas (\(\mathrm{H}_{2}\)) and iodine gas (\(\mathrm{I}_{2}\)).

• The constant \(k\) is the rate constant, which varies with temperature and system specifics.
• The brackets (\([ ]\)) denote molarity, which is concentration in moles per liter.

The rate law tells us about how the rate depends on reactant concentrations, but it does not give us direct information about the reaction mechanism, especially whether it proceeds in a single step or through multiple steps. Sometimes a complex mechanism may still align with the observed rate law.

Activation Energy

Activation energy \( (E_a) \) is the minimum energy required for a reaction to proceed. It is like an energy barrier that reactants must overcome to be converted into products. According to the Arrhenius Equation,\[k = Ae^{\frac{-E_a}{RT}}\]we see that the rate constant \(k\) is influenced by both temperature \((T)\) and activation energy \((E_a)\). Here:

• \(A\) is the pre-exponential factor, which is related to the frequency of collisions and orientation of the molecules.
• \(R\) is the universal gas constant.

Raising the temperature increases the fraction of molecules that have kinetic energy greater than \(E_a\), thereby increasing the rate of reaction. It's important to note that the activation energy is a fixed property of each reaction and does not change with temperature. The exception is when a catalyst is used, as it provides a new pathway with lower activation energy.

Catalysts

Catalysts are substances that speed up reactions without being consumed in the process. They achieve this by providing an alternative reaction pathway with a lower activation energy \((E_a)\). The impact of a catalyst can be understood with the following points:

• They do not change the energies of the products or reactants, so the overall energy change \(\Delta H\) for the reaction remains the same.
• By lowering \(E_a\), more molecules have the necessary energy to react, thus increasing the reaction rate.

Adding a catalyst affects the rate constant \(k\), as it typically increases the proportion of effective collisions, hence accelerating the reaction.

A common example is the use of enzymes in biological systems. Enzymes can catalyze reactions at body temperature, which would otherwise occur very slowly or not at all.

Reaction Order

The reaction order describes how the rate is affected by the concentration of reactants. It is determined by the exponents in the rate law. For the given reaction,\[\text{Rate} = k[\mathrm{H}_{2}][\mathrm{I}_{2}]\]the reaction is first order with respect to each reactant:

• First order in \(\mathrm{H}_2\) because the exponent is 1.
• First order in \(\mathrm{I}_2\) for the same reason.

Thus, the overall reaction order is the sum of the individual orders, which in this case is 2 (1+1).

Understanding reaction orders is crucial for predicting how the rate changes with concentration. For example, doubling the concentration of \(\mathrm{H}_2\) and \(\mathrm{I}_2\) results in a rate increase by the factor of four, consistent with it being a second-order reaction.