Question

The following initial rate data were obtained for the reaction: \[ 2\text{NO (g)} + \text{Br}_2 \text{(g)} \rightarrow 2 \text{NOBr (g)} \]

Hint:

When determining the order of a reaction with respect to reactants, use the method of comparing experiments with different concentrations of one reactant while keeping others constant. This will help isolate the effect of each reactant on the rate.

Answer 1

The rate law for the reaction is given by: \[\text{Rate} = k[\text{NO}]^p[\text{Br}_2]^q\] where \( p \) is the order with respect to NO and \( q \) is the order with respect to $\text{Br}_{2}$. (a) Determining the order with respect to NO (\( p \)) and $\text{Br}_{2}$ (\( q \)). To determine \( p \), compare experiments 1 and 2 where the $\text{Br}_{2}$ concentration is constant. From experiments 1 and 2: \[\frac{1 \times 10^{-3}}{3 \times 10^{-3}} = \frac{k[0.05]^p[0.05]^q}{k[0.05]^p[0.15]^q}\] \[\frac{1}{3} = \left( \frac{0.05}{0.15} \right)^q \Rightarrow \frac{1}{3} = \left( \frac{1}{3} \right)^q \Rightarrow q = 1\] To determine \( q \), compare experiments 1 and 3 where the NO concentration is constant. From experiments 1 and 3: \[\frac{9 \times 10^{-3}}{1 \times 10^{-3}} = \frac{k[0.15]^p[0.05]^q}{k[0.05]^p[0.05]^q}\] \[\frac{9}{1} = \left( \frac{0.15}{0.05} \right)^p \Rightarrow 9 = \left( 3 \right)^p \Rightarrow p = 2\] Thus, the order with respect to NO is 2 and the order with respect to $\text{Br}_{2}$ is 1. (b) Calculating the rate constant (k). Using the rate law and data from experiment 1: \[1 \times 10^{-3} = k[0.05]^2[0.05]\] \[k = \frac{1 \times 10^{-3}}{(0.05)^3} = 8 \, \text{L}^2 \, \text{mol}^{-2} \, \text{s}^{-1}\] (c) Determining the rate when [NO] = 0.4 M and [$\text{Br}_{2}$] = 0.2 M. Using the calculated rate constant and the given concentrations: \[\text{Rate} = k[\text{NO}]^2[\text{Br}_2]\] \[\text{Rate} = 8 \, \text{L}^2 \, \text{mol}^{-2} \, \text{s}^{-1} \times (0.4 \, \text{mol L}^{-1})^2 \times (0.2 \, \text{mol L}^{-1})\] \[\text{Rate} = 8 \times 0.16 \times 0.2 \, \text{mol L}^{-1} \, \text{s}^{-1} = 2.56 \times 10^{-1} \, \text{mol L}^{-1} \, \text{s}^{-1}\]