Question

(a) The reaction \(\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\frac{1}{2} \mathrm{O}_{2}(g)\) is first order. At 300 \(\mathrm{K}\) the rate constant equals \(7.0 \times 10^{-4} \mathrm{s}^{-1}\) . Calculate the half-life at this temperature. (b) If the activation energy for this reaction is \(75 \mathrm{kJ} / \mathrm{mol},\) at what temperature would the reaction rate be doubled?

Short answer

The half-life of the reaction at 300 K is 990 seconds. The reaction rate would be doubled at approximately 332.34 K.

Step-by-step solution

Find the half-life for a first order reaction at 300K

The half-life for a first order reaction is given by the formula: \(t_{1/2} = \frac{0.693}{k}\) Where \(t_{1/2}\) is the half-life and \(k\) is the rate constant. In this case, the rate constant (\(k\)) is given as \(7.0 \times 10^{-4} s^{-1}\). Let's use the formula to find the half-life: \(t_{1/2} = \frac{0.693}{7.0 \times 10^{-4} s^{-1}}\)

Calculate the half-life

Now, plug in the given values and solve for the half-life: \(t_{1/2} = \frac{0.693}{7.0 \times 10^{-4} s^{-1}} \approx 990 s\) So, the half-life of the reaction at 300 K is 990 seconds. Now let's move on to part (b) of the exercise.

Use the Arrhenius equation to find the temperature at which the reaction rate doubles

The Arrhenius equation is given by: \(k = Ae^{-\frac{E_a}{RT}}\) Where: \(k\) is the rate constant \(A\) is the pre-exponential factor (a constant) \(E_a\) is the activation energy \(R\) is the gas constant (\(8.314 J/(mol \cdot K)\)) \(T\) is the temperature in Kelvin Since we want to find the temperature at which the reaction rate doubles, we can set up the equation like this: \(2k = Ae^{-\frac{E_a}{RT'}}\) Here, \(T'\) is the new temperature at which the reaction rate is doubled.

Solve for the new temperature \(T'\)

Now, we can divide the original Arrhenius equation by the new equation to eliminate the constant A: \(\frac{2k}{k}=e^{\frac{E_a}{RT}-\frac{E_a}{RT'}}\) Simplify the left side of the equation: \(2=e^{\frac{E_a}{RT}-\frac{E_a}{RT'}}\) Now, let's take the natural logarithm of both sides: \(\ln{2}=\frac{E_a}{RT}-\frac{E_a}{RT'}\) Next, we can rearrange the equation to find \(T'\), knowing that \(E_a = 75,000J/mol\), \(R = 8.314 J/(mol \cdot K)\), and the initial temperature \(T = 300K\): \(T'=\frac{E_aRT}{E_aR -RT\ln{2}}\) Now, plug in the values: \(T'=\frac{75,000 J/mol \cdot 8.314 J/(mol \cdot K) \cdot 300K}{75,000 J/mol \cdot 8.314 J/(mol \cdot K) -300K \cdot 8.314 J/(mol \cdot K) \cdot \ln{2}}\)

Calculate the new temperature \(T'\)

Now, solve for \(T'\): \(T' \approx 332.34 K\) So, the reaction rate would be doubled at approximately 332.34 K.

Key concepts

Understanding the Rate Constant

When studying chemical kinetics, a pivotal concept is the rate constant, symbolized by k. This value is intrinsic to a reaction and provides a measure of its speed; essentially, it tells us how fast a reaction will proceed under certain conditions. For a first-order reaction, like the decomposition of hydrogen peroxide (\textsf{\textbf{H}}\(_2\)\textsf{\textbf{O}}\(_2\)) in water, the rate constant is particularly important since it directly relates to the reaction's half-life (t_{1/2}).

The half-life is the time taken for the concentration of a reactant to decrease to half its initial concentration and for a first-order reaction, it is calculated using the equation:
\[ t_{1/2} = \frac{0.693}{k} \]
With a higher rate constant, the half-life decreases, indicating a faster reaction. It is essential for students to grasp that the rate constant, although a 'constant', is actually dependent on factors like temperature, which leads us into the Arrhenius equation and how it describes the temperature dependence of k.

The Arrhenius Equation and Temperature's Effect

Delving deeper into chemical kinetics, the Arrhenius equation establishes a quantitative link between the rate constant (\textsf{\textbf{k}}) and temperature. It is expressed as:
\[ k = Ae^{-\frac{E_a}{RT}} \]
where A is the pre-exponential factor, E_a represents the activation energy, R is the universal gas constant, and T is the absolute temperature in Kelvin.

The equation essentially says that the rate constant will increase with a rise in temperature, leading to a faster reaction rate. For instance, if we know the activation energy and the rate constant at a particular temperature, we can predict the change in the rate constant with a change in temperature.

This aspect is vital when considering real-world applications where temperature control is used to optimize reaction rates, such as in the synthesis of chemicals, pharmaceuticals, or even in the biological processes occuring within our own bodies.

Activation Energy and the Reaction Barrier

Integral to the Arrhenius equation is the concept of activation energy (E_a), the minimum energy that must be supplied to reactants for a chemical reaction to take place. It's akin to a barrier that reactants must overcome to transform into products. The higher the activation energy, the slower the reaction rate, as fewer molecules possess the necessary kinetic energy to react at a given temperature.

Given that the activation energy for the hydrogen peroxide decomposition reaction is 75 kJ/mol, we can infer it's relatively high, meaning that under typical conditions, the reaction is slow. However, by increasing temperature, more molecules gain the required energy to overcome this barrier, thereby increasing the reaction rate.

This relationship between activation energy and temperature is crucial in industries and research where altering reaction rates is necessary, and it is beautifully illustrated when calculations show that doubling the reaction rate of our given exercise only requires a slight increase in temperature to 332.34 K, proving how sensitive reaction rates are to temperature alterations.