Problem 28 - Chemistry: Principles and Reactions

Question

Diethylhydrazine reacts with iodine according to the following equation: $$\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2}(\mathrm{NH})_{2}(l)+\mathrm{I}_{2}(a q) \longrightarrow\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{~N}_{2}(l)+2 \mathrm{HI}(a q)$$ The rate of the reaction is followed by monitoring the disappearance of the purple color due to iodine. The following data are obtained at a certain temperature. $$ \begin{array}{cccc} \hline \text { Expt. } & {\left[\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2}(\mathrm{NH})_{2}\right]} & {\left[\mathrm{I}_{2}\right]} & \text { Initial Rate }(\mathrm{mol} / \mathrm{L} \cdot \mathrm{h}) \\ \hline 1 & 0.150 & 0.250 & 1.08 \times 10^{-4} \\ 2 & 0.150 & 0.3620 & 1.56 \times 10^{-4} \\ 3 & 0.200 & 0.400 & 2.30 \times 10^{-4} \\ 4 & 0.300 & 0.400 & 3.44 \times 10^{-4} \\ \hline\end{array}$$ (a) What is the order of the reaction with respect to diethylhydrazine, iodine, and overall? (b) Write the rate expression for the reaction. (c) Calculate \(k\) for the reaction. (d) What must \(\left[\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2}(\mathrm{NH})_{2}\right]\) be so that the rate of the reaction is \(5.00 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{h}\) when \(\left[\mathrm{I}_{2}\right]=0.500 ?\)

Step-by-Step Solution

Answer

Question: Determine the concentration of diethylhydrazine (C2H5)2(NH)2 when the rate of the reaction is 5.00 x 10^-4 mol/L·h and the concentration of iodine (I2) is 0.500 M. Use the given data in the table. Answer: The concentration of diethylhydrazine must be 0.347 M for the rate of the reaction to be 5.00 x 10^-4 mol/L·h when the concentration of iodine is 0.500 M.

Step 1: (a) Finding the order of the reaction

To find the order of the reaction, we can compare the initial rates and concentrations of reactants from different experiments. From experiments 1 and 2, the concentration of diethylhydrazine is the same, so we can compare the initial rates and find the order for iodine. \(\frac{(1.56\times10^{-4})}{(1.08\times10^{-4})}=\left(\frac{[I_2]_2}{[I_2]_1}\right)^m\) \(\frac{(1.56\times10^{-4})}{(1.08\times10^{-4})}=\left(\frac{0.3620}{0.250}\right)^m\) Solving for m, we get: m = 1, which means that the reaction is first order with respect to iodine. Now, let's compare the experiments 3 and 4 to find the order for diethylhydrazine. \(\frac{(3.44\times10^{-4})}{(2.30\times10^{-4})}=\left(\frac{[C_2H_5]_4}{[C_2H_5]_3}\right)^n\) \(\frac{(3.44\times10^{-4})}{(2.30\times10^{-4})}=\left(\frac{0.300}{0.200}\right)^n\) Solving for n, we get: n = 1, which means that the reaction is first order with respect to diethylhydrazine. Since the reaction is first order with respect to both reactants, the overall order of the reaction is 1+1=2.

Step 2: (b) Writing the rate expression for the reaction

Now that we know the order of the reaction with respect to diethylhydrazine and iodine, we can write the rate expression for the reaction: Rate = k [diethylhydrazine]^n [iodine]^m = k[(\(C_2H_5)_2(NH)_2\)][I2]

Step 3: (c) Calculating the rate constant (k)

We can use any of the given experiments to calculate the rate constant (k). Let's use experiment 1 data: \(1.08\times10^{-4}=\)(k)(0.150)(0.250) Solving for k, we get: k = \(2.88\times10^{-3} \mathrm{L/mol·h}\)

Step 4: (d) Finding the concentration of diethylhydrazine

We're given that: Rate=\(5.00\times10^{-4}\mathrm{mol/L·h}\) \([I_2]=0.500\) We need to find the concentration of diethylhydrazine, i.e., [\((C_2H_5)_2(NH)_2\)]. We can use the rate expression for the reaction: \(5.00\times10^{-4}=(2.88\times10^{-3})[_(C_2H_5)2(NH)_2](0.500)\) Solving for [\((C_2H_5)_2(NH)_2\)], we get: [\((C_2H_5)_2(NH)_2\)]=\(\frac{5.00\times10^{-4}}{(2.88\times10^{-3})(0.500)}\) [\((C_2H_5)_2(NH)_2\)]= 0.347 So, the concentration of diethylhydrazine must be 0.347 M for the rate of the reaction to be \(5.00 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{h}\) when \([I_2]=0.500\).

Key Concepts

Understanding Reaction Kinetics

Reaction kinetics is the study of the rates at which chemical processes occur and the factors that affect these rates. It is a key concept in understanding how reactions happen and how to control them. The speed at which reactants turn into products is referred to as the reaction rate, and it can be affected by variables such as temperature, concentration of reactants, and the presence of catalysts.

Monitoring a reaction, like observing the disappearance of the purple color due to iodine in the given exercise, is essential in kinetics. The rate data obtained from experiments at a certain temperature can be used to calculate the order with respect to each reactant, which tells us how the reaction rate depends on the concentration of that particular reactant. In our exercise, experiments were methodically designed to keep one reactant concentration constant while varying another, allowing for calculations of the individual order of reaction for the reactants and the overall order.

Deciphering the Rate Law Expression

The rate law expression is a mathematical formula that describes the relationship between the rate of a reaction and the concentrations of its reactants. It is determined empirically, meaning it is based on experimental data rather than theoretical predictions. This expression includes a rate constant, which embodies the intrinsic reactivity of the reacting species under specific conditions, and exponents which represent the reaction order with respect to each reactant.

In the exercise, we deduced that the reaction was first order in both diethylhydrazine and iodine. Thus, the rate law expression takes the form: Rate = k [(C2H5)2(NH)2][I2]. This simplifies understanding and predicting how changing the concentration of either reactant will affect the reaction rate, allowing for tailor-made changes to the reaction conditions when necessary.

Rate Constant Calculation Procedure

The rate constant, symbolized as 'k', crucially influences the reaction rate and varies with temperature and the presence of catalysts. Calculating 'k' involves using the rate law expression with known rates and concentrations from experimental data.

In the provided example, with the established rate law, we can use any of the experimental data points to solve for 'k'. The process involves simple algebra to isolate and calculate the value of 'k'. Knowing the rate constant allows for prediction of reaction rates under new conditions of concentration, a powerful tool in chemical synthesis and process optimization. Understanding how to calculate and apply 'k' can empower students and chemists to manipulate reactions to their benefit, enabling more efficient and controlled chemical processes.