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Determining Rate Constant

In Rate Equations, we learned that rate of reaction is linked to two things: The concentrations of certain species, and a particular constant, k. If we don't know the value of this constant, it is impossible to work out the rate of a chemical reaction. Determining the rate constant is an important step in writing rate equations, which allow us to…

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Determining Rate Constant

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Jetzt kostenlos anmeldenIn **Rate Equations**, we learned that rate of reaction is linked to two things: The **concentrations of certain species**, and a particular constant, **k**. If we don't know the value of this constant, it is impossible to work out the rate of a chemical reaction. **Determining the rate constant **is an important step in writing rate equations, which allow us to accurately predict the rate of a reaction under certain conditions.

- This article is about
**determining the rate constant**in physical chemistry. - We'll start by
**defining rate constant**. - We'll then consider the
**importance of the rate constant**. - After that, we'll learn how you
**determine the rate constant units**. - Next up, we'll look at two different ways of
**determining the rate constant experimentally**, using**initial rates**and**half-life data**. - You'll be able to have a go at calculating the rate constant yourself with our
**worked examples**. - Finally, we'll take a deep dive into a
**rate constant formula**, which links the rate constant to the**Arrhenius equation**.

The **rate constant**,** k**, is a **proportionality constant **that links the **concentrations of certain species **to the** rate of a chemical reaction**.

Every chemical reaction has its own **rate equation**. This is an expression that can be used to predict the rate of the reaction under specific conditions, provided you know certain details. As we explored in the introduction, the rate equation is linked to both the **concentrations of certain species**, and the r**ate constant**. Here's how they're related:

Note the following:

- k is the
**rate constant**, a value that is constant for each reaction at a particular temperature. We're interested in k today. - The letters A and B represent
**species involved in the reaction**, be they reactants or catalysts. - Square brackets show
**concentration**. - The letters m and n represent
**the order of the reaction with respect to a particular species**. This is the power that the species' concentration is raised to in the rate equation. - Overall, [A]
^{m}represents the**concentration of A, raised to the power of m**. This means that it has the**order of m**.

Species involved in the rate equation tend to be reactants but they can also be catalysts. Likewise, not every reactant is necessarily part of the rate equation. For example, take a look at the following reaction:

$$I_2+CH_3COCH_3\rightarrow CH_3COCH_2I+HI$$

Its rate equation is given below:

$$\text{rate} =k[H^+][CH_3COCH_3]$$

Note that H^{+} *does *appear in the rate equation, despite not being one of the reactants. On the other hand, reactant I_{2} *doesn't* appear in the rate equation. This means that the concentration of I_{2} has no effect on the rate of reaction whatsoever. This is the definition of a zeroth order reaction.

Let's take a moment to consider why the rate constant matters so much in chemistry. Suppose you had a reaction with the following rate equation:

$$\text{rate} =k[A][B]$$

What if the value of our rate constant was extremely large - say, 1 × 10^{9}? Even if we had very low concentrations of A and B, the rate of reaction would still be pretty fast. For example, if our concentrations of A and B were just 0.01 mol dm^{-3} each, we'd get the following rate of reaction:

$$\begin{align} \text{rate} &=(1\times 10^9)(0.01)(0.01)\\ \\ \text{rate} &=1\times 10^5\space mol\space dm^{-3}\space s^{-1}\end{align}$$

That's certainly not to be laughed at!

But on the other hand, what if the value of our rate constant was extremely small - how about 1 × 10^{-9}? Even if we had very high concentrations of A and B, the rate of reaction wouldn't be fast at all. For example, if our concentrations of A and B were 100 mol dm^{-3} each, we'd get the following rate of reaction:

$$\begin{align} \text{rate} &=(1\times 10^{-9})(100)(100)\\ \\ \text{rate} &=1\times 10^{-5}\space mol\space dm^{-3}\space s^{-1}\end{align}$$

That's very slow!

A **large rate constant** means that the rate of reaction is likely to be **fast**, even if you use low concentrations of the reactants. But a **small rate constant** means that the rate of reaction is likely to be **slow**, even if you use large concentrations of reactants.

In conclusion, the rate constant plays an important role in dictating the **rate of a chemical reaction**. It gives scientists another way of influencing the rate of a reaction beyond simply changing concentrations, and can dramatically increase the profitability of industrial processes.

Before we learn how to determine the rate constant, k, we need to find out how to **determine its units**. Provided you know the rate equation, the process is simple. Here are the steps:

- Rearrange the rate equation to make k the subject.
- Substitute the units of concentration and rate of reaction into the rate equation.
- Cancel the units through until you are left with the units of k.

Here's an example. We'll then use it to determine the rate constant in the next part of this article.

**A reaction has the following rate equation:**

$$\text{rate} =k[A][B]^2$$

**Concentration and rate are given in mol dm ^{-3} and mol dm^{-3} s^{-1} respectively. Calculate the units of k.**

To solve this problem, we first rearrange the rate equation given in the question to make k the subject:

$$k=\frac{\text{rate}}{[A][B]^2}$$

We then substitute the units for rate and concentration, also given in the question, into this equation:

$$k=\frac{mol\space dm^{-3}\space s^{-1}}{(mol\space dm^{-3})(mol\space dm^{-3})^2}$$

We can then expand the brackets and cancel the units down to find the units of k:

$$\begin{align} k&=\frac{mol\space dm^{-3}\space s^{-1}}{mol^3\space dm^{-9}}\\ \\ k&=mol^{-2}\space dm^6\space s^{-1}\end{align}$$

That is our final answer.

For all you mathematicians out there, we have a much quicker way of working out the units of the rate constantIt involves using the overall order of the reaction. All reactions with the same order, no matter how many species they include, end up having the same units for their rate constant.

Let's look at that more closely.

Consider a second-order reaction. It could have either of these two rate equations:

$$\text{rate} =k[A][B]\qquad \qquad \text{rate} =k[A]^2$$

But in rate equations, concentration always has the same units: mol dm^{-3}. If we rearrange the two expressions to find the units of k using the method we describe above, they both end up looking the same:

$$\begin{gather} k=\frac{mol\space dm^{-3}\space s^{-1}}{(mol\space dm^{-3})(mol\space dm^{-3})}\qquad \qquad k=\frac{mol\space dm^{-3}\space s^{-1}}{(mol\space dm^{-3})^2}\end{gather}$$ $$k=mol^{-1}\space dm^3\space s^{-1} $$

We can extrapolate these results to come up with a general formula for the units of k, where n is the order of the reaction:

$$k=\frac{mol\space dm^{-3}\space s^{-1}}{(mol\space dm^{-3})^n}$$

If it suits you, you could simplify the fraction even further using **exponential rules**:

$$k=mol^{1-n}\space dm^{-3+3n}\space s^{-1}$$

**Work out the units of k for a generic first-order reaction.**

We could find the units of k in either of two ways: Using the fraction, or using the simplified formula. It doesn't matter which method we choose - we'll end up getting the same answer. Here, the reaction is first-order and so n = 1. In both cases, the units of k simplify down to just s^{-1}.

$$\begin{gather} k=\frac{mol\space dm^{-3}\space s^{-1}}{(mol\space dm^{-3})^1}\qquad \qquad k=mol^{1-1}\space dm^{-3+3}\space s^{-1}\\ \\ k=mol^0\space dm^0\space s^{-1}\\k=s^{-1}\end{gather}$$

We've now reached the main focus of this article: **Determining the rate constant**. We'll look in particular at **determining the rate constant ****through experimental methods**.

To find the rate equation, and so to be able to confidently predict the rate of a reaction, we need to know the **order of the reaction with respect to each species**, as well as the **rate constant**. If you want to learn how to find out the *order of a reaction*, check out **Determining Reaction Order**, but if you instead wish to learn how to calculate the *rate constant*, stick around - this article has got you covered.

We'll focus on two different methods:

- Initial rates.
- Half-life data.

First up - calculating the rate constant from **initial rates of reaction**.

One way of getting enough information to calculate the rate constant is through **initial rates data**. In **Determining Reaction Order**, you learned how you can use this technique to find the order of the reaction with respect to each species. We'll now take the process one step further and use the orders of reaction we worked out to calculate the rate constant.

Here's a reminder of how you use initial rates data to find the order of reaction with respect to each species.

- Carry out the same chemical reaction experiment again and again, keeping almost all the conditions the same each time, but varying the concentrations of reactants and catalysts.
- Plot a concentration-time graph for each reaction and use the graph to find each experiment's
**initial rate**. - Mathematically compare the initial rates with the different concentrations of species used to find the order of the reaction with respect to each species, and write these into the rate equation.

You're now ready to use the orders of reaction to find the rate constant k. Here are the steps you should take:

- Choose one of the experiments.
- Substitute the values of concentration used and the initial rate of reaction determined for that particular experiment into the rate equation.
- Rearrange the equation to make k the subject.
- Solve the equation to find the value of k.
- Find the units of k as described earlier in the article.

Let's show you how. We'll then use the rate equation in its entirety to calculate the rate of the same reaction, but using different concentrations of species.

**You carry out experiments in class and end up with the following initial rates data:**

[A] (mol dm^{-3}) | [B] (mol dm^{-3}) | Rate of reaction (mol dm^{-3} s^{-1}) | |

Reaction 1 | 1.0 | 1.0 | 0.5 |

Reaction 2 | 2.0 | 1.0 | 1.0 |

**The value of the rate constant, k.****The initial rate of reaction under the same conditions, using 1.16****mol dm**^{-3}of A and 1.53**mol dm**^{-3}of B.

First, let's find k. We can use what we are told about the orders of the reaction with respect to both A and B to write a rate equation.

$$\text{rate} =k[A][B]^2$$

Note that we looked at this rate equation earlier on in the article, and so we already know the units that k will take: mol^{-2} dm^{6} s^{-1}.

For the next step, we need to use data from one of the experiments. It doesn't matter which experiment we choose - they should all give us the same answer for k. We simply substitute the concentrations of A and B used in the experiment, as well as the initial rate of reaction, into the rate equation. We then rearrange it slightly, solve the equation, and end up with a value for k.

Let's take reaction 2. Here, the rate of reaction is 1.0 mol dm^{-3} s^{-1}, the concentration of A is 2.0 mol dm^{-3}, and the concentration of B is 1.0 mol dm^{-3}. If we put these values into the rate equation given, we get the following:

$$1.0 =k(2.0)(1.0)$$

We can rearrange the equation to find the value of k.

$$\begin{gather} k=\frac{1.0}{(2.0)(1.0)^2}=\frac{1.0}{2.0}\\ \\ k=0.5\space mol^{-2}\space dm^6\space s^{-1}\end{gather}$$

That's the first part of the question done. The second part wants us to predict the initial rate of reaction for the same reaction but using different concentrations of A and B. We do this by substituting the concentrations that the question gives us, alongside our calculated value of k, into the rate equation. Remember that the units of rate of reaction are mol dm^{-3} s^{-1}.

$$\begin{gather} \text{rate} =k[A][B]^2\\ \\ \text{rate} =0.5(1.16)(1.53)^2\\ \\ \text{rate} =1.36mol^{-2}\space dm^6\space s^{-1}\end{gather}$$

This is our final answer.

**Half-lives** offer us another way of determining the rate constant, k. You might know from **Determining Reaction Order** that the **half-life ****(t**_{1/2}**)** of a species is the time it takes for half of the species to be used in the reaction. In other words, it is the time it takes for its **concentration to halve**.

There are a few interesting things about half-life when it comes to rate equations. First, if the half-life of a species is **constant** throughout the reaction, no matter its concentration, then you know that the reaction is **first order** in respect to that species. But half-life also relates numerically to the **rate constant** with certain formulae. The formula depends on the overall order of the reaction. For example, **if the reaction itself is first-order**, then the rate constant and the half-life of the reaction are linked in the following way:

$$k=\frac{\ln(2)}{t_{1/2}}$$

You'll find different equations linking half-life and the rate constant for reactions with different orders. Check with *your* exam board to find out which formulae you need to learn.

Let's break the equation down:

- k is the rate constant. For first-order reactions, it is measured in s
^{-1}. - ln(2) means the logarithm of 2, to the base e. It is a way of asking, "if e
^{x}= 2, what is x?" - t
_{1 /2 }is the half-life of the first-order reaction, measured in seconds.

Using half-life to find the rate constant is simple:

- Convert the half-life of the reaction into seconds.
- Substitute this value into the equation.
- Solve to find k.

Here's an example to help you understand how the process is done.

**A sample of hydrogen peroxide has a half-life of 2 hours. It decomposes in a first-order reaction. Calculate the rate constant, k, for this reaction.**

To calculate k, we first need to convert the half-life, which is 2 hours, into seconds:

$$2\times 60\times 60=7200\space s$$

We then simply substitute this value into the equation:

$$\begin{gather} k=\frac{\ln(2)}{7200}\\ \\ k=9.6\times 10^{-5}\space s^{-1}\end{gather}$$

Remember that we found out the units of the rate constant for all first-order reactions earlier in the article.

You might also see rate constant calculations using **integrated rate laws**. Integrated rate laws relate the concentration of species involved in the rate equation at certain points in the reaction to the rate constant. Their general form differs depending on the order of the reaction.

Integrated rate laws are typically used once you know the rate equation and rate constant to calculate how long it will take to reduce the concentration of a species to a particular level. However, we can do the opposite - provided we know the order of the reaction and have information about concentrations at different points in the reaction, we can calculate the rate constant.

Sound complicated? Don't worry - you don't need to know how to work with integrated rate laws at A level. But if you plan to study chemistry at a higher level, you might find it interesting to get ahead and read all about them. Try asking your teacher for any recommended resources to kick-start your learning.

Lastly, let's consider another formula for the rate constant. It relates the rate constant, k, to the Arrhenius equation:

Here's what that all means:

- k is the
**rate constant**. Its units vary depending on the reaction. - A is the
**Arrhenius constant**, also known as the pre-exponential factor. Its units also vary, but are always the same as the rate constant's. - e is
**Euler's number**, approximately equal to 2.71828. - E
_{a}is the**activation energy**of the reaction, with the units J mol^{-1}. - R is the
**gas constant**, 8.31 J K^{-1}mol^{-1}. - T is the
**temperature**, in K. - Overall, \(e^\frac{-E_a}{RT} \) is the proportion of molecules that have enough energy to react.

If you want to see some examples of the equation in action, or fancy practising calculating the rate constant from the Arrhenius equation, check out **Arrhenius Equation Calculations**.

Here's a question - can you come up with a range of values that the rate constant k always falls in? For example, can k ever be negative? Could it equal zero?

To answer this question, let's use the Arrhenius equation:

$$k=Ae^\frac{-E_a}{RT} $$

For k to be negative, either A or \(e^\frac{-E_a}{RT} \) must be negative. Likewise, for k to equal exactly zero, either A or \(e^\frac{-E_a}{RT} \) must equal exactly zero. Is this possible?

Well, exponentials are **always greater than zero***.* They might get very close to zero, but they never quite reach it, and so they are always positive. Try using a scientific calculator online to raise e to the power of a large negative number, such as -1000. You'll get an *infinitesimally** small* value - but it will still be positive. For example:

$$e^{-1000}=3.72\times 10^{-44}$$

That number is still above zero!

So, \(e^\frac{-E_a}{RT} \) can't be negative or equal to zero. But can A?

If you've read **Arrhenius Equation**, you'll know that A is the **Arrhenius constant**. To simplify the subject down, A is all to do with the number and frequency of collisions between particles. Particles are always moving, and so they are always colliding. In fact, particles would only stop moving if we reached absolute zero, which is energetically impossible! Therefore, A is **always greater than zero**.

Well, we've learned that both A and \(e^\frac{-E_a}{RT} \) must always be greater than zero. They are always positive, and can't be negative or exactly equal to zero. Therefore, k must also always be positive. We can summarise this mathematically:

$$\begin{gather} A\gt 0\qquad e^\frac{-E_a}{RT}\gt 0\\ \\ \therefore k\gt 0 \end{gather}$$

We're at the end of this article. By now, you should understand what we mean by the **rate constant** and why it is important in chemical reactions. You should also be able to** determine the units of the rate constant** using the **rate equation**. In addition, you should feel confident **calculating the rate constant **using **initial rates** and **half-life data**. Finally, you should know the formula that links the **rate constant and the Arrhenius equation**.

- The
**rate constant**,**k**, is a**proportionality constant**that links the**concentrations of certain species**to the**rate of a chemical reaction**. - A
**large rate constant**contributes to a**fast rate of reaction**, whilst a**small rate constant**often results in a**slow rate of reaction**. - We
**determine the units of the rate constant**using the following steps:- Rearrange the rate equation to make k the subject.
- Substitute the units of concentration and rate of reaction into the rate equation.
- Cancel the units through until you are left with the units of k.

We can

**determine the rate constant experimentally**using**initial rates**or**half-life data**.To calculate the rate constant using

**initial rates**:- Substitute experimental values of concentration and rate of reaction into the rate equation.
- Rearrange the equation to make k the subject and solve to find k.

- To calculate the rate constant using
**half-life**:- Convert the half-life of the reaction into seconds.
- Substitute this value into the equation and solve to find k.

- The rate constant relates to the
**Arrhenius equation**with the formula \(k=Ae^\frac{-E_a}{RT} \)

To find the rate constant for any reaction, you can use the rate equation and initial rates data. However, to find the rate constant of a first-order reaction in particular, you can also use half-life. The half-life of a first-order reaction (t_{1/2}) and the reaction's rate constant are linked using a particular equation: k = ln(2) / t_{1/2}

Alternatively, you can find the rate constant using integrated rate laws. However, this knowledge goes beyond A level content.

More about Determining Rate Constant

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