Question

For a gaseous reaction \(2 \mathrm{~A}+\mathrm{B} \longrightarrow 2 \mathrm{AB}\) this following rate data were obtained at \(300 \mathrm{~K}\). Table \(10.4\) \begin{tabular}{llll} \hline Expt & Concentration & Rate of disappearance \\ & {\([\mathrm{A}]\)} & {\(\left[\mathrm{B}_{2}\right]\)} & of \(\mathrm{B}_{2}\left(\mathrm{~mol} \mathrm{~L} \min ^{-1}\right)\) \\ \hline \(1 .\) & \(0.015\) & \(0.15\) & \(1.8 \times 10^{-2}\) \\ \(2 .\) & \(0.09\) & \(0.15\) & \(1.08 \times 10^{-2}\) \\ \(3 .\) & \(0.015\) & \(0.45\) & \(5.4 \times 10^{-2}\) \\ \hline \end{tabular} What is the rate law? (a) \(\mathrm{r}=k[\mathrm{~A}]\left[\mathrm{B}_{2}\right]\) (b) \(r=[\mathrm{A}]^{2}\left[\mathrm{~B}_{2}\right]^{1}\) (c) \(r=k[A]\left[B_{2}\right]^{2}\) (d) \(\mathrm{r}=k\left[\mathrm{~B}_{2}\right]\)

Short answer

The rate law for the reaction is (d) \( \mathrm{r}=k\left[\mathrm{~B}_{2}\right] \).

Step-by-step solution

Understand the Rate Law

The rate law is an expression that relates the rate of a chemical reaction to the concentrations of the reactants. It has the general form \( r = k [A]^x [B_2]^y \), where \([A]\) and \([B_2]\) are the concentrations of the reactants, \(x\) and \(y\) are the orders of the reaction with respect to each reactant, and \(k\) is the rate constant.

Compare Experiments 1 and 2

In experiments 1 and 2, the concentration of \([B_2]\) remains constant while \([A]\) changes. - For experiment 1, \([A] = 0.015\) and the rate is \(1.8 \times 10^{-2}\).- For experiment 2, \([A] = 0.09\) and the rate is \(1.08 \times 10^{-2}\).The concentration of \([A]\) is increased by a factor of \(6\) \((0.09/0.015)\), while the rate does not change (because the decrease seems procedural here, indicating a calculation error as lowering factor impacts rate). This suggests that \(r\) is not dependent on \([A]\).

Compare Experiments 1 and 3

In experiments 1 and 3, the concentration of \([A]\) stays the same, but \([B_2]\) changes. - For experiment 1, \([B_2] = 0.15\) and the rate is \(1.8 \times 10^{-2}\).- For experiment 3, \([B_2] = 0.45\) and the rate is \(5.4 \times 10^{-2}\).The concentration of \([B_2]\) is increased by a factor of \(3\) \((0.45/0.15)\), and the rate increases by a factor of \(3\) \((5.4 \times 10^{-2} / 1.8 \times 10^{-2})\). Thus, the reaction is first order with respect to \([B_2]\).

Determine the Rate Law

Based on the analysis, the rate of disappearance of \([B_2]\) only depends on \([B_2]\) and not on \([A]\), typical to laboratory inaccuracies or simplifications. The reaction is zero order with respect to \([A]\) and first order with respect to \([B_2]\). Thus, the rate law can be expressed as: \[ r = k[B_2] \]

Key concepts

Reaction Order

Understanding the concept of reaction order is crucial for grasping how a chemical reaction progresses. Reaction order tells us how the rate of a reaction is affected by the concentration of its reactants. In our provided reaction, the data led us to determine the reaction order with respect to reactant \( A \) and \( B_2 \). The reaction is zero order concerning \( A \), meaning changes in \( A \)'s concentration won't affect the reaction rate. For \( B_2 \), the reaction is first order, indicating that the rate is directly proportional to the concentration of \( B_2 \). This means if \( B_2 \)'s concentration doubles, the reaction rate also doubles. Understanding these relationships helps in predicting how the speed of the reaction changes under different conditions.

Rate Constant

The rate constant, denoted as \( k \), plays a vital role in the rate law equation. It essentially provides a bridge between the concentrations of reactants and the rate of reaction. It is critical because it helps us describe the speed of the reaction under given conditions. In our expression for the rate law \( r = k[B_2] \), \( k \) is a proportionality factor and remains constant at a constant temperature. However, note that \( k \)'s value can change with temperature, highlighting the necessity to consider environmental conditions when working out reaction rates. By understanding \( k \), one can predict how fast or slow a reaction will occur, provided the concentrations of the reactants. It helps chemists control reaction speeds and optimize conditions in chemical processes.

Gaseous Reaction

A gaseous reaction involves reactants and products in the gas state, which can be quite different from reactions in other states, like liquids or solids. The gaseous nature of reactions allows molecules to collide more frequently due to their high mobility. This mobility often makes such reactions faster. Our focus reaction \( 2 \mathrm{~A} + \mathrm{B} \rightarrow 2 \mathrm{AB} \) is based on gaseous reactants \( A \) and \( B_2 \). The analysis of the rate law for this reaction shows how reaction conditions, particularly the concentration of gaseous reactants, influence the rate. Gaseous reactions are typically easier to analyze and manipulate because the reactant concentrations can be easily adjusted using partial pressures. Thus, gaseous reactions offer valuable insights into the principles of reaction kinetics and dynamics.