Problem 49 - Chemistry: Principles and Reactions

Question

The decomposition of \(\mathrm{Y}\) is a zero-order reaction. Its half-life at \(25^{\circ} \mathrm{C}\) and \(0.188 M\) is 315 minutes. (a) What is the rate constant for the decomposition of Y? (b) How long will it take to decompose a \(0.219 \mathrm{M}\) solution of \(\mathrm{Y}\) ? (c) What is the rate of the decomposition of \(0.188 \mathrm{M}\) at \(25^{\circ} \mathrm{C}\) ? (d) Does the rate change when the concentration of \(\mathrm{Y}\) is increased to \(0.289 \mathrm{M}\) ? If so, what is the new rate?

Step-by-Step Solution

Answer

Question: For the given zero-order decomposition reaction of Y, find (a) the rate constant at a concentration of 0.188 M and a half-life of 315 minutes, (b) the time it takes for a 0.219 M solution of Y to decompose, (c) the rate of decomposition at a concentration of 0.188 M and a temperature of 25°C, and (d) the new rate when the concentration of Y is increased to 0.289 M. Answer: (a) The rate constant (k) for the decomposition reaction of Y is 2.98 x 10^-4 M/min. (b) It will take approximately 736 minutes for the 0.219 M solution of Y to decompose. (c) The rate of decomposition at a concentration of 0.188 M and a temperature of 25°C is 2.98 x 10^-4 M/min. (d) The new rate when the concentration of Y is increased to 0.289 M remains 2.98 x 10^-4 M/min, as the rate is independent of the reactant concentration in a zero-order reaction.

Step 1: (a) Finding the rate constant

For a zero-order reaction, the integrated rate law equation is given by: $$\left[ Y \right] = -kt + \left[ Y \right]_{0}$$ We are given the half-life (t_1/2) of the reaction at a specific concentration and temperature. The half-life relationship for a zero-order reaction is: $$t_1/2 = \frac{\left[ Y \right]_0}{2k}$$ We can rearrange this equation to find the rate constant (k): $$k = \frac{\left[ Y \right]_0}{2t_1/2}$$ Plugging in the given values: $$k = \frac{0.188 \mathrm{M}}{2(315 \: \mathrm{min})}$$ #Calculation of k

Step 2: Now, we can calculate the rate constant (k): $$k = \frac{0.188 \mathrm{M}}{630 \: \mathrm{min}} = 2.98 \times 10^{-4} \: \mathrm{M/min}$$ This is the rate constant for the decomposition of Y.

(b) Time to decompose 0.219 M solution

Step 3: To find the time it takes for the 0.219 M solution to decompose, we can use the integrated rate law equation for a zero-order reaction: $$\left[ Y \right] = -kt + \left[ Y \right]_{0}$$ Rearrange the equation to find the time (t): $$t = \frac{\left[ Y \right]_{0} - \left[ Y \right]}{k}$$ Plugging in the given values: $$t = \frac{0.219 \mathrm{M} - 0 \mathrm{M}}{2.98 \times 10^{-4} \: \mathrm{M/min}}$$ #Calculation of time

Now, we can calculate the time: $$t = \frac{0.219 \mathrm{M}}{2.98 \times 10^{-4} \: \mathrm{M/min}} = 735.57 \: \mathrm{min}$$ It will take approximately 736 minutes for the 0.219 M solution of Y to decompose.

Step 4: (c) Rate of decomposition

For a zero-order reaction, the rate is given by: $$rate = k\left[ Y \right]^0 = k$$ So at a concentration of 0.188 M at 25°C, the rate of decomposition is equal to the rate constant: $$rate = 2.98 \times 10^{-4} \: \mathrm{M/min}$$

Step 5: (d) New rate when the concentration of Y increases

Since this is a zero-order reaction, the rate is independent of the reactant concentration. Therefore, increasing the concentration of Y from 0.188 M to 0.289 M does not change the rate. The new rate remains: $$rate = 2.98 \times 10^{-4} \: \mathrm{M/min}$$

Key Concepts

Rate Constant Calculation

The rate constant, often denoted by the symbol 'k', is a pivotal factor in the study of reaction kinetics. It provides a direct measure of how quickly a reaction proceeds under given conditions. For zero-order reactions, the rate constant is particularly easy to calculate once you have the half-life and the initial concentration.

Here's a simplified explanation: the half-life of a zero-order reaction (the time it takes for half of the reactant to be consumed) is directly proportional to the initial concentration. The formula is \[ k = \frac{\left[ Y \right]_0}{2t_{1/2}} \], where \(\left[ Y \right]_0\) is the initial concentration and \(t_{1/2}\) is the half-life. By inserting the given values into the equation, we compute 'k' which remains constant regardless of the concentration of the reactant Y — a distinctive feature of zero-order kinetics.

It's important to understand that this constant value of 'k' is only valid at a specific temperature; as temperature changes, so does 'k'. Furthermore, in real-world scenarios, concentrations may not be uniform throughout the system, so 'k' provides an average rate over time and space.

Reaction Half-Life

The concept of half-life is critical when discussing reaction kinetics, as it provides a tangible measure of the reaction's progression over time. For a zero-order reaction, the half-life is quite straightforward: it's the time it takes for the concentration of the reactant to reduce by half.

The formula to determine the half-life of a zero-order reaction is derived from the integrated rate law and is given by \[ t_{1/2} = \frac{\left[ Y \right]_0}{2k} \], where \(\left[ Y \right]_0\) is the initial concentration and 'k' is the rate constant. This equation shows the inverse relationship between half-life and initial concentration: as the initial concentration increases, the half-life decreases for a zero-order reaction.

Understanding half-life can be particularly useful in predicting how much time it will take for a reactant to decrease to a certain level, which is crucial in fields such as pharmacology, where the half-life of a drug in the body influences dosing schedules.

Integrated Rate Law

The integrated rate law for zero-order reactions is the tool that connects concentration, time, and rate constant. It is a mathematical expression that enables the calculation of reactant concentration at any given time or the time required for the reactant concentration to reach a particular level.

The zero-order integrated rate law is \[ \left[ Y \right] = -kt + \left[ Y \right]_{0} \], signifying that the change in concentration over time is linearly related to time, not exponentially as with first-order reactions. In other words, the rate at which the reactant is used up remains constant over time.

The practicality of this rate law becomes clear when we rearrange it to solve for time, given by \[ t = \frac{\left[ Y \right]_{0} - \left[ Y \right]}{k} \]. This is useful for planning reactions in a lab setting or controlling industrial processes. Remember, since the rate of reaction is zero-order, it remains steady, unaffected by shifts in the reactant's concentration, provided that temperature and other conditions are unchanged.