Question

Complete the following table for the reaction \(2 \mathrm{R}(g)+3 \mathrm{~S}(g) \longrightarrow\) products that is first-order in \(\mathrm{R}\) and second-order in \(\mathrm{S}\) $$ \begin{array}{lcccc} \hline & & \boldsymbol{k} & \text { Rate } \\ & {[\mathrm{R}]} & {[\mathrm{S}]} & \left(\mathrm{L}^{2} / \mathrm{mol}^{2} \cdot \mathrm{min}\right) & (\mathrm{mol} / \mathrm{L} \cdot \mathrm{min}) \\ \hline \text { (a) } & 0.200 & 0.200 & 1.49 & \\ \text { (b) } & & 0.633 & 0.42 & 0.833 \\ \text { (c) } & 0.100 & & 0.298 & 0.162 \\ \text { (d) } & 0.0500 & 0.0911 & & 0.00624 \\ \hline \end{array} $$

Short answer

Question: Calculate the missing values in the given table for the reaction \(2 \mathrm{R}(g) + 3 \mathrm{S}(g) \longrightarrow\) products, which is first-order in R and second-order in S. Answer: The missing values in the table are as follows: (b) row: [R] = 1.152 mol/L (c) row: [S] = 0.234 mol/L (d) row: Rate = 0.003079 mol/L.min

Step-by-step solution

Calculate missing values in the concentration of R and S

Using the rate law equation, we will calculate the missing concentrations of R and S. For (b) row, we have: `rate = k[R][S]^2` 0.833 = 0.42[R](0.633)^2 Calculate the concentration of R: [R] = 1.152 For (c) row, we have: `rate = k[R][S]^2` 0.162 = 0.298(0.1)[S]^2 Calculate the concentration of S: [S] = 0.234

Calculate the missing Rate value for (d) row

Using the rate law equation, we will find the rate for the reaction (d). Rate law: `rate = k[R][S]^2` For (d) row, we have: Rate = 0.0500 * (0.0911)^2 * (k) Given k for (a) row: k = 1.49 (L^2/mol^2.min) Calculate the Rate: Rate = (0.0500) * (0.0911)^2 * (1.49) = 0.003079 (mol/L.min)

Fill in the missing values in the table

Now we can complete the table with the calculated values in Step 1 and Step 2. $$ \begin{array}{lcccc} \hline \\ & \boldsymbol{k} & \text { Rate } \\ & {[\mathrm{R}]} & {[\mathrm{S}]} & \left(\mathrm{L}^{2} / \mathrm{mol}^{2} \cdot \mathrm{min}\right) & (\mathrm{mol} / \mathrm{L} \cdot \mathrm{min}) \\ \hline \text { (a) } & 0.200 & 0.200 & 1.49 & \\ \text { (b) } & 1.152 & 0.633 & 0.42 & 0.833 \\ \text { (c) } & 0.100 & 0.234 & 0.298 & 0.162 \\ \text { (d) } & 0.0500 & 0.0911 & 1.49 & 0.003079 \\ \hline \end{array} $$

Key concepts

Rate Law Equation

Understanding the rate law equation is fundamental when studying the speed of chemical reactions. The rate law expresses the relationship between the rate of a chemical reaction and the concentration of its reactants. For a general reaction where reactants A and B form products, its rate can be represented as rate = k[A]^x[B]^y, where k is the rate constant, [A] and [B] indicate the molar concentrations of the reactants, and x and y are the reaction orders with respect to A and B, respectively. The units of k vary depending on the overall reaction order, which is the sum of individual orders.

Using the given exercise as an example, we consider a reaction that is first-order in R and second-order in S. We can write the rate law as rate = k[R][S]^2. To find the rate or missing reactant concentrations, we rearrange this equation, plugging in known values and solving for the unknown. It's like solving a puzzle where each piece is a piece of data from the experiment, allowing students to find the missing information and complete the table.

Reaction Order

Reacting to 'reaction order' requires an understanding of how it impacts the rate of a chemical reaction. The reaction order indicates the power to which the concentration of a reactant is raised in the rate law equation. It tells us how the rate changes with a change in concentration of that reactant. If the reaction order is 1, then the rate is directly proportional to the reactant concentration. If it's 2, the rate changes with the square of the concentration changes, and so on. This concept helps in figuring out the complex dynamics of how chemicals interact.

In our exercise, reaction order reveals the specific reliance on concentrations of R and S. An order of 1 for R and 2 for S means as R's concentration changes, the rate changes at the same pace, whereas for S, the rate shifts at a pace squared to the change in concentration. This understanding is crucial when predicting how the reaction rate will vary with changes in concentration during an experiment or industrial process.

Chemical Kinetics

Diving deep into chemical reactions requires an appreciation for chemical kinetics, the study of reaction rates and how they change under different conditions. This area of chemistry explores the speed at which reactants transform into products and the factors influencing this speed—including concentration, which we focus on in rate laws, but also temperature, catalysts, and even the physical state of the reactants.

Knowledge of kinetics is more than academic; it's practical. Understanding it helps chemists to design and optimize reactions, control product formation or the breakdown of harmful substances in the environment. For instance, budding chemists looking at our example exercise can observe practical kinetics in action. They must use their knowledge of rate laws and reaction orders, inherent to chemical kinetics, to solve for unknowns in a tabulated format, much like how real-world kinetic data is analyzed in laboratories and industries worldwide.