Chemistry, 14.07.2019 10:00 bhenley

Diamond is forever" is one of the most successful advertising slogans of all time. but is it true? for the reaction shown below, calculate the standard free energy change at 298k and determine whether

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Answer from: mochoa4

The given reaction is

$C _{(Diamond)}\rightarrow C_{(Graphite)}$

An element can exist in 2 or more different forms which have totally different chemical and physical properties. They are known an allotropes.

Diamond and Graphite are allotropic forms of carbon. In the given reaction, diamond is changing to graphite and we have to find out the standard free energy of this reaction. This will help us to find out whether this reaction is spontaneous at 298 K or not.

The following data is needed for calculations which is taken from standard reference table.

$H_{f}^{0}(diamond)= 1.895 kJ/mol$

$H_{f}^{0}(graphite)= 0$

$S^{0}(diamond)=2.337 J/mol-K$

$S^{0}(graphite)=5.740 J/mol-K$

Step 1: Find ΔH⁰ rxn for the given reaction.

The formula to calculate ΔH⁰ rxn is given below.

$\bigtriangleup H^{0}_{rxn}= H_{f}(product)- H_{f}( reactant)$

We have graphite on product side and diamond on reactant side.

Therefore, $\bigtriangleup H^{0}_{rxn}= H_{f}(graphite)- H_{f}(diamond)$

Let us plug in the values given above.

$\bigtriangleup H^{0}_{rxn}= 0 - 1.895 kJ/mol$

$\bigtriangleup H^{0}_{rxn}= - 1.895 kJ/mol$

ΔH⁰ rxn for the given reaction is -1.895 kJ/mol

Step 2 : Find ΔS⁰ rxn for the given reaction.

The formula to calculate ΔS⁰ rxn is

$\bigtriangleup S^{0}_{rxn}= S^{0}(product)- S^{0}( reactant)$

$\bigtriangleup S^{0}_{rxn}= S^{0}(graphite)- S^{0}(diamond)$

$\bigtriangleup S^{0}_{rxn}= (5.740J/mol.K) - (2.337 J/mol.K)$

$\bigtriangleup S^{0}_{rxn}= 3.403 J/mol.K$

Let us convert this to kJ.

$\frac{3.403J}{mol.K}\times \frac{1 kJ}{1000J}= 3.403\times 10^{-3}kJ/mol.K$

ΔS⁰ rxn for the given reaction is 3.403 x 10⁻³ kJ/mol.K

Step 3: Find standard free energy ΔG⁰ rxn.

ΔG⁰ rxn for the given reaction is calculated as

$\bigtriangleup G^{0}_{rxn}= \bigtriangleup H^{0}_{rxn}- T\times \bigtriangleup S^{0}_{rxn}$

We have T = 298 K. Let us plug in the calculated values of ΔH⁰ rxn and ΔS⁰ rxn.

$\bigtriangleup G^{0}_{rxn}= - 1.895 kJ/mol - [(298K)\times 3.403\times 10^{-3}kJ/mol.K]$

$\bigtriangleup G^{0}_{rxn}= - 1.895 kJ/mol - [1.014 kJ/mol]$

$\bigtriangleup G^{0}_{rxn}= - 2.909 kJ/mol$

The standard free energy change for the given reaction is -2.909 kJ/mol

The negative value of delta G⁰ suggests that the given reaction is spontaneous at room temperature. That means diamond will slowly convert to graphite. The speed of this reaction is extremely slow, but yet the reaction is taking place. So over a period of time diamond will become graphite.

Therefore "Diamond is forever" is not true as it is going to get converted to graphite.

Answer from: Quest

if you mean hydrogen and oxygen then water is the product

Answer from: Quest

If a hydrogen atom is neutral, it must also have one electron. an oxygen atom (o) has an atomic number of eight (8). this means a neutral oxygen atom has eight protons and eight electrons. the element actium (ac) has an atomic number of 89, so it has 89 protons and 89 electrons in a neutral atom