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

Rate law for the reaction can be expressed as: \[ \text{Rate} = k[\text{NO}]^p[\text{Br}_2]^q \] 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 (\( p \)) Compare experiments 1 and 2 to eliminate the effect of $\text{Br}_{2}$ concentration and find \( p \). From experiments 1 and 2: \[ \frac{1 \times 10^{-3}}{3 \times 10^{-3}} = \frac{k[0.05]^p[0.05]^q}{k[0.05]^p[0.15]^q} \] \[ \frac{1}{3} = \left( \frac{1}{3} \right)^q \quad \Rightarrow \quad q = 1 \] Determining the order with respect to $\text{Br}_{2}$ (\( q \)) Compare experiments 1 and 3 to eliminate the effect of NO concentration and find \( p \). From experiments 1 and 3: \[ \frac{9 \times 10^{-3}}{1 \times 10^{-3}} = \frac{k[0.15]^p[0.05]^q}{k[0.05]^p[0.05]^q} \] \[ \frac{9}{1} = \left( \frac{3}{1} \right)^p \quad \Rightarrow \quad p = 2 \] Thus, the order with respect to NO is 2 and the order with respect to $\text{Br}_{2}$s is 1. (b) Calculating the rate constant (k) Use the rate law and the data from any experiment to solve for \( k \). Using 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 the concentrations of NO and $\text{Br}_{2}$ are 0.4 M and 0.2 M Now that we have the value of \( k \), we can use it to calculate the rate when [NO] = 0.4 M and [$\text{Br}_{2}$] = 0.2 M. \[ \text{Rate} = k[\text{NO}]^2[\text{Br}_2] \] \[ \text{Rate} = 8 \, \text{L}^2 \, \text{mol}^{-2} \, \text{s}^{-1} \times (0.4)^2 \times 0.2 \] \[ \text{Rate} = 8 \times 0.16 \times 0.2 = 2.56 \times 10^{-1} \, \text{mol L}^{-1} \, \text{s}^{-1} \]