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  Standard enthalpy of formation, Chemistry tutorial

Introduction:

Define: The standard enthalpy of formation is stated as the change in enthalpy whenever one mole of a substance in the standard state (that is, 1 atm of pressure and 298.15 K) is made up from its pure elements under the similar conditions.

The standard enthalpy of formation is the measurement of the energy consumed or released when one mole of a substance is formed under standard conditions from its pure elements. The symbol of the standard enthalpy of formation is ΔHof.

Δ = A change in enthalpy

o = A degree signifies that it is a standard enthalpy change

f = The 'f' points out that the substance is made up from its elements

The equation for standard enthalpy change of formation (originating from the Enthalpy's being a State Function), illustrated below, is generally used:

ΔHoreaction = ∑ ΔHof (products) - ∑ΔHof (Reactants)

This equation necessarily defines that the standard enthalpy change of formation is equivalent to the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants.

Concepts of Standard Enthalpy of Formation, ΔHof

The standard enthalpy of formation (ΔHof) is stated as the enthalpy change for making one mole of a compound from elements in their standard states. The standard states are the form we would find out an element in at 1 bar pressure and 25 °C.

For illustration, the standard enthalpy of formation for CH4 would be represented as ΔHof (CH4 (g)) and would be the enthalpy change for the given reaction.

C (s, graphite) + 2H2 (g) → CH4 (g)

Carbon is represented as being graphite as the energy would be dissimilar for the absurd reaction run with the diamond form of carbon. As hydrogen is a molecular gas under standard conditions, this is the form needed in the standard reaction.

This is significant to note that the enthalpy of formation for any element in its standard form will be accurately zero as for such a reaction the reactants (that is, initial state) and the products (that is, the final state) are similar and therefore can't encompass any change in enthalpy. For illustration: the reaction for the formation of molecular oxygen gas is:

O2 (g) → O2 (g), ΔH = 0

The enthalpy of formation of the non-standard form is non-zero.  As well, the enthalpy of formation will based on the state of the compound which is being formed.  For illustration, the enthalpy of formation of gaseous water will be more than the enthalpy of formation of liquid water (that is, they will be different by the enthalpy of vaporization of water, ΔHvap).

Enthalpies (or heats) of formation are very helpful in computing the reaction enthalpies. That is as any reaction can be visualized as taking place through a path in which first all the reactant compounds are transformed to elements and then all the elements are transformed in the product compounds.  The very first step in this method will be negative of the sums of the enthalpies of formation of the reactants.  Negative as it is the reverse of the formation reaction (that is, compounds to elements). The secondary step is the sum of the enthalpies of formation of the products.

ΔHor = ΣnΔH°f (prod) - ΣnΔH°f (react)

Calculating the Standard Enthalpy of Reaction:

Compute the standard enthalpy of reaction for the combustion of methane:

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)   ΔHorxn =?

In order to compute the standard enthalpy of reaction, we require looking up the standard enthalpies of formation for each of the reactants and products comprised in the reaction. These are generally found in an appendix or in different tables online. For this reaction, the data we require is:

ΔHof {CH4 (g)} = - 75kJ/mol

ΔH of {O2 (g)} = 0kJ/mol

ΔH of {CO2 (g)} = - 394kJ/mol

ΔH of {H2O (g)} = - 284kJ/mol

It will be noted that as it exists in its standard state, the standard enthalpy of formation for oxygen gas is 0 kJ/mol. After that, we sum up our standard enthalpies of formation. Remember that as the units are in kJ/mol, we require multiplying by the stoichiometric coefficients in the balanced reaction equation.

∑ΔH of {products} = ΔH of {CO2 (g)} + ΔH of {H2O (g)}

= (1) (- 394) + (2) (- 284) = - 962 kJ/mol

∑ΔH of {reactants} = ΔH of {CH4 (g)} + ΔH of {O2 (g)}

= (1) (- 75) + (2) (0) = - 75kJ/mol

Now, we can determine the standard enthalpy of reaction:

ΔHorxn = ∑ ΔHof {products} - ∑ ΔHof {reactants}

= (- 962) - (- 75) = - 887kJ/mol

As we would anticipate, the standard enthalpy for this combustion reaction is strongly exothermic.

Key theories for doing enthalpy computations:

1) Whenever a reaction is reversed, the magnitude of ΔH stays similar, however the sign changes.

2) If the balanced equation for a reaction is multiplied via an integer, the corresponding value of ΔH should be multiplied by that integer as well.

3) The change in enthalpy for a reaction can be computed from the enthalpies of formation of the reactants and products.

4) Elements in their standard states make no contribution to the enthalpy computations for the reaction as the enthalpy of the element in its standard state is zero.

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