Feasibility Of Conversions Through Analysis Of Enthalpy Changes

The overall heat-content of a system is called Enthalpy. Enthalpy change is the amount of heat absorbed or released in a reaction at constant pressure. Symbol of enthalpy change is delta H. When heat is absorbed the enthalpy change is positive. When heat is released the enthalpy change is negative.

Standard enthalpy change is the change in enthalpy when one mole of a substance is formed from its pure elements in the standard state. Standard state is 1 atmospheric of pressure and 298.15 kelvin of temperature. Standard enthalpy change of a reaction is equal to the sum of standard enthalpy of formation of products minus sum of standard enthalpy of formation of reactants.

Let us calculate the enthalpy change in a process. Methanol is converted into formaldehyde and hydrogen gas. Enthalpy change during this process can be calculated by following way.First of all we need to find the heat of combustion of methanol, formaldehyde and hydrogen gas which are given in illustration.We shall now determine the enthalpy change for the conversion of methanol into formaldehyde and hydrogen gas.

First of all we shall rearrange these equations according to our main reaction. Methanol is on the reactant side so we shall write combustion of methanol equation as it is.One mole of methanol is taken in our main reaction but in combustion reaction there is moles of methanol. We shall divide the whole methanol combustion equation by two. We shall also divide heat of combustion by two as shown in the illustration.

Because formaldehyde is on the product side in main reaction but in combustion reaction it is in reactant side. We shall turn around the combustion reaction of formaldehyde. When turning around the equation, we shall also turn around the plus-minus-sign of heat of reaction. As we can see in the given illustration there is one mole of formaldehyde in main reaction and also in combustion reaction.

Now for the combustion of hydrogen we can see there is hydrogen gas on product side in main reaction and in reactant side in the combustion. We shall turn around the combustion equation along with the heat of combustion.After that we shall divide the combustion equation by two because there is one mole of hydrogen gas in main equation and two moles of hydrogen gas in combustion equation.

After arrangement of combustion reactions we shall add up these reactions and their heat of combustion to get final reaction and change in enthalpy value.We can see in the illustration that the change in enthalpy for the conversion of methanol into formaldehyde and hydrogen gas is +86 kJ.

Enthalpy diagram represents the change in enthalpy of a reaction. A reaction can be exothermic or endothermic.On x-axis process of reaction is given. On y-axis change in enthalpy is given.Now lets calculate change in enthalpy for the formation of benzene from carbon and hydrogen.Now we shall write the combustion reactions for reactants and products in our main reaction and then draw a cycle to calculate enthalpy change.

Heat of combustion of one mole of carbon is -394 kJ and for one mole of hydrogen it is -286 kJ. As there are six moles of carbon and three moles of hydrogen in reactant so we have multiplied heat of combustions by six and three and added them together.Now if we observe that the delta H2 is -3268kJ. We shall turn around the equation and plus-minus-sign of heat of combustion to complete the cycle towards product.

Now we shall add the delta H2 and delta H3 to get the enthalpy change for our reaction. We shall also draw the enthalpy cycle.Enthalpy cycle also shows that delta H3 process was exothermic and delta H2 process was endothermic. When we added both of these values we get the enthalpy change value for our reaction. The enthalpy change for the formation of one mole of benzene from carbon and hydrogen is +46 kJ.

We can measure standard enthalpy of dissolution experimentally. First of all take a plastic cup and add some solvent in it and cover it completely. Now put the thermometer in it and measure the initial temperature. After that put a weighed quantity of solute in solvent and stir it until it is completely dissolved. Now measure the final temperature. After that measure the change in temperature. Now put values of mass, specific heat capacity of solvent and temperature in the formula given in illustration and calculate heat q of the solution. We can find the mass of solvent by multiplying its density with volume used. It can be exothermic or endothermic reaction.Now find the number of moles of solute by dividing the mass of solute by molar mass of solute.Finally, divide the value of heat of solution by number of moles of solute taken. The resulting answer is standard heat of dissolution of that solute.

Enthalpy of neutralization is the amount of heat absorbed or released when one mole of water is formed by the reaction of aqueous solution of acid and aqueous solution of base under standard conditions.We can experimentally determine the enthalpy of neutralization. First of all take a Styrofoam cup and put aqueous solution of base in it. Now put thermometer in it and measure the initial temperature. Cover the cup completely.Now put solution of acid in it. Then measure the final temperature. Calculate the change in temperature. Now put values of mass of solution specific heat capacity and change in temperature in the formula given in illustration and find heat released. Finally find number of moles of the water formed from neutralization reaction. Divide heat of the solution by number of moles of water formed. The final resulting value will be enthalpy of neutralization.

When a more electropositive metal displaces one mole of metal from its salt solution under standard conditions then change in enthalpy occurs. This change in enthalpy is called enthalpy of displacement.When we react zinc metal with a copper sulphate solution, copper is displaced by zinc metal. Enthalpy change during this process is -210 kJ/mol. This is enthalpy of displacement.

We shall now experimentally determine the enthalpy of displacement of one mole of copper ions in copper sulphate solution by magnesium metal. We take a plastic cup. Then we put the thermometer in it. After that we put fifty centimeter cube of zero point one molar copper sulphate solution in it and measure the initial temperature for two minutes. We note-down the initial temperature. Then we put two gram of magnesium powder in it. We stir the mixture and after some time note-down the final temperature.

Now we note-down the initial and final temperature values and measure the change in the temperature.After that we find the number of moles of copper from number of moles of copper sulphate solution used. We then find the change in enthalpy by formula delta H equals mass multiplied with heat capacity c multiplied with theta. Theta represents the change in temperature.

We get the change in enthalpy for displacement of zero point zero one mole of copper. We finally find the change in enthalpy for displacement of one mole of copper. All calculations are shown in given illustration.

When one mole of a compound is formed from its pure elements under standard conditions, then enthalpy change occurs. This enthalpy change is called standard enthalpy of formation. It is denoted by ∆Hf.When writing the equation for standard enthalpy of formation we should always write one mole of compound in product side. If it requires to write coefficient in fraction on reactant side, that is completely fine.

Let us take an example of enthalpy of formation of water. One mole of water is formed from hydrogen gas and oxygen gas. Energy is released in this process. Negative two hundred and ninety six kilo joule per mole of heat is released in this process.This is an exothermic process which means heat is released in the process.This heat is called standard enthalpy of formation.

In another example carbon reacts with oxygen to form the benzene. As we can see one mole of benzene is formed. Energy is absorbed in this process.This absorbed energy is called enthalpy of formation.

When we burn one mole of compound under standard conditions in oxygen then there is a change in enthalpy. This enthalpy change is called standard enthalpy of combustion. In the given illustration we are carrying out the combustion of one mole of hydrogen in presence of oxygen to form water. The standard enthalpy of combustion of this reaction is negative two hundred and eighty six kilo joule per mole.

In another example we are burning ethane. Standard enthalpy of combustion of ethane is -1560kJ/mol.When writing the equation for enthalpy of combustion we write compound to be burnt as one mole no matter if oxygen is written in fractions. Moreover, the reactants and products are written in their standard state. For example, water exists as liquid in standard state. So water is written as liquid (l).

When ethanol burns in the presence of oxygen then carbon dioxide and water is formed. -1366.8 kJ/mol of heat is released in the process. This heat is called standard enthalpy of combustion. The reaction takes place under standard conditions of 1 atmospheric of pressure and 298.15 kelvin of temperature.

The energy stored in bonds between atoms in a molecule is called bond enthalpy. Enthalpy of bond dissociation is the amount of enthalpy that is added for the homolytic cleavage of the bond between two atoms in a molecule.In the given illustration the bond between the molecule AB is broken symmetrically. This results in the formation of radical.The enthalpy is required to break this bond.

Four hundred and thirty six kilo joule per mole of enthalpy is required to break the bond symmetrically between two hydrogen atoms in a hydrogen molecule.This enthalpy required to break the bond is called bond dissociation enthalpy.After the breakage of bond the hydrogen radicals are formed.

In Prop-1-ene the double bond is broken symmetrically between two carbon atoms. The enthalpy required to break this bond is six hundred and ten kilo joule per mole.Radicals are formed as a result.This enthalpy is called enthalpy of bond dissociation.

When we react solutions of acid and solution of base to produce one mole of water under standard conditions then enthalpy change occurs. This enthalpy change is called standard enthalpy of neutralization.When an acid reacts with a base it forms salt and water. This reaction is called neutralization reaction.We react sodium hydroxide in aqueous form with hydrochloric acid in aqueous state to form aqueous sodium chloride and one mole of water. The enthalpy change in this process is -57.9 kJ/mol.This is termed as standard enthalpy of neutralization. Heat is released in this process.

The amount of energy released or absorbed when one mole of solute is dissolved in water under standard conditions is called standard enthalpy of solvation.When one mole of solid aluminium chloride is dissolved in water, -373.63 kJ of heat is released. The standard enthalpy of solvation of aluminium chloride is -373.63 kJ/mol.

The enthalpy change that occurs when we dissolve one mole of gaseous ions in sufficient amount of water to form infinite dilute solution under standard conditions is called Standard enthalpy of hydration.Infinite dilute solution means there will be no change in enthalpy if we further dilute the solution by adding water.It is denoted by ∆Hhyd.In the given illustration M+ is in gaseous state and after adding water it is in aqueous state.

Enthalpy change for Na+ ion is -405 kJ/mol.Sodium ions in gaseous state are dissolved in water. Then they are in aqueous state. Simillarly lithium ions in gaseous state are dissolved in sufficient water to form infinite dilution. Standard enthalpy of hydration for lithium ions is -520 kJ/mol.

Standard enthalpy of dissolution is the amount of energy or heat released or absorbed when one mole of solute is completely dissolved in solvent under standard conditions. It is represented by ∆Hsoln. It is measured in kJ/mol.It can be calculated by dividing heat of the solution by number of moles of solute dissolved. Enthalpy of dissolution of ammonium chloride is +14.78 kJ/mol. It means 14.78 kilo joule of heat is absorbed when one mole of ammonium chloride is completely dissolved in a solvent.

Hess’s law states that if a reaction can be carried out in a series of steps then sum of enthalpies of each step should equal the enthalpy change for overall reaction.We can validate this law by first measuring the enthalpy for the conversion of carbon and hydrogen into ethyne gas. This reaction can be written in a series of steps. Enthalpy of each step is measured experimentally by using calorimeter. We also have experimentally measured enthalpy value for our main reaction which is +224 kJ/mol. But we shall find the change in enthalpy of main reaction using Hess’s law. We shall now add the enthalpies of individual reactions. Final enthalpy after adding enthalpies of each reaction step is +226.7kJ which is a value quite near to the experimentally measured value of our main reaction. This validates the Hess’s law.