Hydrogen bromide is a highly reactive and corrosive gas used mainly as a catalyst for organic reactions. It is produced by reacting hydrogen and bromine gases together. $$\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \longrightarrow 2 \mathrm{HBr}(g)$$ The rate is followed by measuring the intensity of the orange color of the bromine gas. The following data are obtained: $$\begin{array}{cccc}\hline \text { Expt. } & {\left[\mathrm{H}_{2}\right]} & {\left[\mathrm{Br}_{2}\right]} & \text { Initial Rate }(\mathrm{mol} / \mathrm{L} \cdot \mathrm{s}) \\ \hline 1 & 0.100 & 0.100 & 4.74 \times 10^{-3} \\ 2 & 0.100 & 0.200 & 6.71 \times 10^{-3} \\ 3 & 0.250 & 0.200 & 1.68 \times 10^{-2} \\ \hline\end{array}$$(a) What is the order of the reaction with respect to hydrogen, bromine, and overall? (b) Write the rate expression for the reaction. (c) Calculate \(k\) for the reaction. What are the units for \(k ?\) (d) When \(\left[\mathrm{H}_{2}\right]=0.455 \mathrm{M}\) and \(\left[\mathrm{Br}_{2}\right]=0.215 M\), what is the rate of the reaction?

Step by step solution

01

Determine the order of the reaction with respect to hydrogen

To determine the order of the reaction with respect to hydrogen, compare Experiment 2 and Experiment 3. Notice that the ratio between [H₂] in Experiment 3 to [H₂] in Experiment 2 is 0.25/0.1 = 2.5. The initial rate of Experiment 3 to Experiment 2 is (1.68 × 10⁻²) / (6.71 × 10⁻³) = 2.5. Since the ratio of initial concentrations equals the ratio of initial rates for hydrogen, we can conclude that the order of the reaction with respect to hydrogen (n) is 1 (first order).

02

Determine the order of the reaction with respect to bromine

To determine the order of the reaction with respect to bromine, compare Experiment 1 and Experiment 2. The ratio between the concentrations of [Br₂] in Experiment 2 to [Br₂] in Experiment 1 is 0.2/0.1 = 2. The initial rate of Experiment 2 to Experiment 1 is (6.71 × 10⁻³) / (4.74 × 10⁻³) ≈ 1.4. Since the ratio of initial concentrations does not equal the ratio of initial rates for bromine, we can conclude that the order of the reaction with respect to bromine (m) is not 1. Try the square root: Taking the square root of the ratio of initial concentrations, we get √(0.2/0.1) = 1.4. Now the ratio of the initial rate equals the square root of the ratio of concentrations. We can conclude that the order of the reaction with respect to bromine is 0.5 (half order).

03

Determine the overall order of the reaction

The overall order of the reaction is the sum of the orders of hydrogen and bromine. Thus, the overall order is 1 (for hydrogen) + 0.5 (for bromine) = 1.5.

04

Write the rate expression for the reaction

Using the reaction order information we obtained in steps 1 and 2, we can write the rate expression for the reaction as follows: $$\text{Rate} = k[\mathrm{H}_{2}]^{1}[\mathrm{Br}_{2}]^{0.5}$$

05

Calculate the rate constant (k) and its units

We can use any experiment data to calculate the rate constant (k). Let's use Experiment 1. Rearrange the rate expression to solve for k, then plug in the given concentrations and rate: $$k = \frac{\text{Rate}}{[\mathrm{H}_{2}]^{1}[\mathrm{Br}_{2}]^{0.5}}$$ $$k = \frac{4.74 \times 10^{-3} M \cdot s^{-1}}{0.1^1 M \cdot (0.1)^{0.5} M^{0.5}}$$ $$k ≈ 0.474\, M^{0.5}s^{-1} $$ So the rate constant, k, is approximately 0.474 M^(0.5)s^(-1).

06

Calculate the rate of the reaction for given concentrations

We are given the concentrations [H₂] = 0.455 M and [Br₂] = 0.215 M, and we need to find the rate of the reaction. We will use the rate expression from Step 4 and substitute the values with the calculated k: $$\text{Rate} = (0.474\,M^{0.5}s^{-1})[(0.455\,M)^{1}][(0.215\,M)^{0.5}]$$ $$\text{Rate} ≈ 0.195\, M\cdot s^{-1}$$ The rate of the reaction, when [H₂] = 0.455 M and [Br₂] = 0.215 M, is approximately 0.195 M/s.

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