Problem 22 - Chemistry: Principles and Reactions

Question

The reaction $$\mathrm{ICl}(g)+\frac{1}{2} \mathrm{H}_{2}(g) \longrightarrow \frac{1}{2} \mathrm{I}_{2}(g)+\mathrm{HCl}(g)$$ is first-order in both reactants. The rate of the reaction is \(4.89 \times 10^{-5} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) when the ICl concentration is \(0.100 M\) and that of the hydrogen gas is \(0.030 \mathrm{M}\) (a) What is the value of \(k\) ? (b) At what concentration of hydrogen is the rate \(5.00 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) and \([\mathrm{ICl}]=0.233 \mathrm{M?}\) (c) At what concentration of iodine chloride is the rate \(0.0934 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) if the hydrogen concentration is three times that of ICl?

Step-by-Step Solution

Answer

(a) The rate constant \(k\) for the given chemical reaction is \(1.63 \times 10^{-3}\ \mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\). (b) When the rate is \(5.00 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) and the concentration of ICl is \(0.233 M\), the concentration of hydrogen gas is \(1.321 M\). (c) When the rate is \(0.0934 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) and the hydrogen concentration is three times that of ICl, the concentration of ICl is \(2.341 M\).

Step 1: Write the rate law equation

Since the reaction is first-order with respect to both reactants, the rate law equation is given by: $$rate = k[\mathrm{ICl}][\mathrm{H}_{2}]$$

Step 2: Use the given information to find the value of \(k\)

We are given the rate as \(4.89 \times 10^{-5} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) when the concentration of ICl is \([\mathrm{ICl}] = 0.100 M\) and the concentration of hydrogen gas is \([\mathrm{H}_{2}] = 0.030 M\). Substituting these values into the rate law equation, we get: $$4.89 \times 10^{-5} = k(0.100)(0.030)$$ Now solve for \(k\): $$k = \frac{4.89 \times 10^{-5}}{(0.100)(0.030)} = 1.63 \times 10^{-3}\ \mathrm{L} / \mathrm{mol} \cdot \mathrm{s}$$ (b) Find the concentration of hydrogen gas when the rate is \(5.00 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) and the concentration of ICl is \(0.233 M\)

Step 3: Replace the known values into the rate law equation

We are given the rate, the value of \(k\) from part (a), and the concentration of ICl. We need to find the concentration of hydrogen gas \([\mathrm{H}_{2}]\). Use the rate law equation: $$5.00 \times 10^{-4} = (1.63 \times 10^{-3})(0.233)[\mathrm{H}_{2}]$$

Step 4: Solve for the concentration of hydrogen gas

Rearrange the equation to solve for \([\mathrm{H}_{2}]\): $$[\mathrm{H}_{2}] = \frac{5.00 \times 10^{-4}}{(1.63 \times 10^{-3})(0.233)} = 1.321 \mathrm{M}$$ So the concentration of hydrogen gas at this rate is \(1.321 M\). (c) Find the concentration of ICl when the rate is \(0.0934 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) and the hydrogen concentration is three times that of ICl

Step 5: Write the rate law equation with given conditions

We are given the rate and the relationship between the concentrations of ICl and hydrogen gas. Let \(x\) be the concentration of ICl. Then, the concentration of hydrogen gas is \(3x\). Use the rate law equation: $$0.0934 = (1.63 \times 10^{-3})(x)(3x)$$

Step 6: Solve for the concentration of ICl

Rearrange the equation and solve for \(x\): $$x^2 = \frac{0.0934}{(1.63 \times 10^{-3})(3)} \Rightarrow x = \sqrt{\frac{0.0934}{(1.63 \times 10^{-3})(3)}} = 2.341 \mathrm{M}$$ So the concentration of iodine chloride is \(2.341 M\) under these conditions.

Key Concepts

Rate Law

The rate law describes how the rate of a chemical reaction depends on the concentration of reactants. For a reaction, it is expressed in the form of an equation:

• rate = k[A]^m[B]ⁿ

where k is the rate constant, [A] and [B] are the concentrations of reactants A and B, and m and n are the reaction orders with respect to each reactant respectively.

In our specific reaction involving ICl and H₂ gases, the rate law is derived as:

• rate = k[ICl][H₂]

This indicates that the rate of reaction is directly proportional to the concentrations of both reactants.
Understanding the rate law is essential for predicting how changes in concentration affect the reaction rate. It informs us of the stoichiometry of the rate-determining step of the reaction mechanism.

Reaction Order

Reaction order is the sum of the powers of the concentration terms in the rate law equation and shows how the concentration of each reactant affects the overall reaction rate.

• If a reaction is first-order with respect to a reactant, the rate changes linearly with concentration changes of the reactant.
• For second-order, the rate changes quadratically with the concentration.

In our example reaction, the order is first for both ICl and H₂, making the overall order of the reaction second-order (1+1=2).
Knowing the reaction order helps chemists understand the dynamics of a reaction and how it can be influenced or controlled by altering conditions.

Rate Constant

The rate constant, k, is a critical part of the rate law equation. It provides specific insights into the reaction rate at a given temperature.
For a reaction:

• rate = k[ICl][H₂]

the value of k can be found by taking known values of rate and concentrations and solving for k.
In the exercise provided, using the rate 4.89 × 10^-5 mol/L⋅s, [ICl] = 0.100 M, and [H₂] = 0.030 M, we calculate:

• k = 1.63 × 10^-3 L/mol⋅s

The rate constant is unique to each reaction and depends on factors like temperature. It doesn’t change with concentration changes and serves as an identifier for the reaction’s speed.

Concentration

Concentration pertains to the amount of a substance present in a certain volume. In the context of chemical kinetics, changes in the concentration of reactants can significantly impact the rate of a reaction.
For the example reaction of ICl and H₂, different concentration scenarios were analyzed.
Using given values, such as 0.100 M for ICl and 0.030 M for H₂, allows you to calculate the reaction rate.

• Higher reactant concentrations generally lead to an increase in reaction rate.
• An understanding of concentration aids in controlling the speed of chemical reactions, by adjusting the inputs accordingly.

Concentration directly integrates into the rate law, making it a pivotal parameter in determining how quickly or slowly a reaction proceeds.

Rate of Reaction

The rate of reaction indicates how quickly or slowly reactants are transformed into products in a chemical reaction. It is typically measured in terms of concentration changes over time.
Practically, the rate of reaction is influenced by several factors including:

• Concentration of reactants
• Temperature
• Presence of catalysts
• Surface area of reactants

In the given exercise, the reaction rate is exemplified by specific numeric values, such as 4.89 × 10^-5 mol/L⋅s, which are calculated using the rate law equation and given concentrations.
The rate is not static and changes with varying conditions, a trait that can be exploited to optimize reactions in chemical industries and laboratories.