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?

Short Answer

Expert verified
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 by step solution

01

(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
02

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
03

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.
04

(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: $$rate = 2.98 \times 10^{-4} \: \mathrm{M/min}$$
05

(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: $$rate = 2.98 \times 10^{-4} \: \mathrm{M/min}$$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The chirping rate of a cricket \(\mathrm{X}\), in chirps per minute near room temperature, is $$\mathrm{X}=7.2 t-32$$ where \(t\) is the temperature in \({ }^{\circ} \mathrm{C}\). (a) Calculate the chirping rates at \(25^{\circ} \mathrm{C}\) and \(35^{\circ} \mathrm{C}\). (b) Use your answers in (a) to estimate the activation energy for the chirping. (c) What is the percentage increase for a \(10^{\circ} \mathrm{C}\) rise in temperature?

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?

The decomposition of ammonia on tungsten at \(1100^{\circ} \mathrm{C}\) is zero- order with a rate constant of \(2.5 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{min} .\) (a) Write the rate expression. (b) Calculate the rate when \(\left[\mathrm{NH}_{3}\right]=0.075 M\). (c) At what concentration of ammonia is the rate equal to the rate constant?

If the activation energy of a reaction is \(4.86 \mathrm{~kJ}\), then what is the percent increase in the rate constant of a reaction when the temperature is increased from \(45^{\circ}\) to \(75^{\circ} \mathrm{C}\) ?

Cold-blooded animals decrease their body temperature in cold weather to match that of their environment. The activation energy of a certain reaction in a cold-blooded animal is \(65 \mathrm{~kJ} / \mathrm{mol} .\) By what percentage is the rate of the reaction decreased if the body temperature of the animal drops from \(35^{\circ} \mathrm{C}\) to \(22^{\circ} \mathrm{C} ?\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free