what is activation energy

Activation Energy Arrhenius Equation Determination of Ea Problem Solution What Is Activation Energy ?

Description: This passage is mainly about what is activation energy. In this passage, the writer shows us how to determine the energy of activation of a chemical reaction using the aeneas equation.

In this article, I am attempting to show you how to determine the energy of activation of a chemical reaction using our aeneas equation, the assumption is that you have already studied the rate law, you understand the order of a reaction, you can predict the factors that affect the rate of reaction, you are familiar with the equation for a straight line y is equal to MX plus B.

You are capable of determining the slope of a line, and you also have some understanding of natural logarithms and logarithms to the base 10 and their properties for calculation, we will make a graphical representation of a chemical reaction in terms of the energy changes involved in it, and this is how the graph will look on the y axis.

I have the energy terms on the x axis, it is also called the reaction coordinate or the progress of reaction, the energy of the reactants is around 60 kilo joules, the energy of the products is about 30 kilo joules, if you use Hess’s law, you can determine the enthalpy change for the reaction which is Delta H is equals to sum of the enthalpies of the products minus and sum of the enthalpies of the reactants.

In this case the value is minus 30 kilo joules, it also tells you that this reaction is an exothermic reaction, now the reactants in this case, if you mix them, they will not spontaneously change into products, they will remain as reactants under the given conditions, but you can make the reactants change into products.

If you provide the right quantity of energy, therefore the reactants are prevented from changing into the products by what we call an energy barrier, and the number that represents the energy is called energy, in this case of threshold energies hundred kilo joules, so the reactants have to gain energy so that the energy of the reactants will become 100 kilojoules.

Then if they undergo proper collision, which means that they have the right amount of energy and proper orientation, they may lead to the formation of products in order to calculate the energy of activation, we need the number threshold energy for threshold energy and the energy of the reactants.

If you subtract the energy of the reactants from the threshold energy, we will get the energy of activation for the forward reaction, the energy of activation will be 100 minus 60, which is 40 kilo joules and this value is representing the energy of activation for the forward reaction, so when the molecules gain 40 kilo joules of energy per mol the change in the products, because the collisions at this point will become effective.

We can also calculate the energy of activation for the backward reaction, if you are looking at a reversible reaction, if you want to change the energy of the products back into reactants, you still have to provide threshold energy, and in this case the value of threshold energy for the backward reaction is going to be 40 plus 30, 70 kilo joules which means that the energy of activation for the backward reaction is more than the energy of activation for the forward reaction.

It is easier to change the reactants to products, you can also say the products are more stable than the reactants, now here what I am attempting to do is to determine the value of energy of activation using the Irenaeus equation, so this is the background in terms of how the energy can be represented for a single step reaction.

Iranian equation written as K is equals to a into e raised to minus EA by RT K represents the rate, constant is this, it is determined experimentally a rep represents frequency factor is characteristic for a given reaction, EA is the energy of activation, R is the universal gas constant 8.314 joules per Kelvin per mole, temperature is always measured in Kelvin.

If you are given degrees Celsius, you may want to convert it to a Kelvin, before you solve it, add 273 point 1 4 2 degrees Celsius, you should get the Kelvin, there are two methods for determining energy of activation, the first one that we are going to look at is the graphical method here, we assume that you have a number of values for the different rate constants at different temperatures, which also implies that the rate constant changes with temperature.

We are going to use the equation K is equals to a into e raised to minus EA by RT which is the rainiest equation, now you want to remove the exponent, for that we are going to use the logarithmic function or more precisely the natural logarithms lon K is equal to lon a minus EA by RT.

So here we have removed the exponent e and this equation, now it can be compared to an equation for a straight line y is equals to MX plus B, y is represented by lon K, M is the slope of the line which is represented by EA and R, and X is 1 over T, so the Irenaeus equation is now compared to an equation for a straight line.

If you plot a graph of y versus x and if it’s a straight line, you can determine the slope, that’s what we’re going to do y-axis lon K obtain from the value of y x axis one over t obtained from the value of x, we are getting a straight line here, and you can determine the slope of this line using the equation.

Here the rice can be calculated from the two numbers fell on K, and run can be calculated from the two temperatures 1 over T and the different time intervals, and that will give you the slope of the line EA by are the numerical value of EA by r, if it’s multiplied by the gas constant, it will give you the value of energy of activation in joules.

So if you have different values for K and the temperature at which the K has been determined, you can obtain the energy of activation by drawing a graph similar to this, the next method is to solve the Iranian to determine the energy of activation for two different temperatures, the equation is already known, we are going to rearrange it after taking natural logarithms.

So this equation is the same as the equation that we have seen a moment ago, lon k1 is equal to lon a minus EA by RT and we’re going to call it equation one, so k1 is the rate constant at temperature t1, the same experiment is going to be performed at temperature t2, and we will write the equation, such as this lon k2 or k2 would be the rate constant at a temperature of T 2 and the equation will be equation 2.

Now if you subtract equation 1 from Equation 2 and rearrange them, this is the resulting equation lon K 2 by K 1 is equal to minus EA by r into 1 by t2 minus 1 by t1, this can be further simplified and you will get this final equation, you can use this equation to determine the energy of activation provided, you’re given the rate constants K 2 K 1 T 1 and T 2.

The value of R will be given to you as it is a constant, so this is the equation that you’re going to learn and you’re going to use for solving for energy of activation, here is a sample problem for you, we are attempting to determine the energy of activation, if the temperature of the reaction system is increased by increased from 300 Kelvin to 310 Kelvin, the rate of the reaction is doubled value of RS 8.314 joules per Kelvin per mole.

So here you’re given the two temperatures, t1 is 300 Kelvin, t2 is 310 Kelvin, it also states that the rate of the reaction doubles when the temperature increases by 10 degrees, which means that the ratio of K 2 by K 1 is going to be to language that is very important here, so that’s the implied meaning.

So if I say the rate of the reaction doubles, the ratio K 2 by K 1 is going to be 2 R is given, we have to calculate energy of activation and we know the equation to pick, so once you have the equation, you’re going to substitute the values K 2 by K k1 is 2, therefore lon 2 into R 8.314 joules per Kelvin mole times 300 into 3 10 divided by 3 10 minus 300 Kelvin lon 2 is 0.6931, therefore the final value of energy of activation will be 5 3 5 9 4 joules or 53.59 4 kilojoules.

Now you can use the same equation and even determine the K value for a different temperature provided, you know the energy of activation, if you like the article, please don’t forget to rate comment and subscribe, thank you and have a great day.

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