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zavuch27 [327]
3 years ago
6

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
Chemistry
1 answer:
Zielflug [23.3K]3 years ago
7 0

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.

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How do you find the amount of moles is .032 grams of water and whats the answer
masha68 [24]

Answer:

\boxed {\boxed {\sf 0.0018 \ mol \ H_2 O }}

Explanation:

First, we need to find the molecular mass of water (H₂O).

H₂O has:

  • 2 Hydrogen atoms (subscript of 2)
  • 1 Oxygen atom (implied subscript of 1)

Use the Periodic Table to find the mass of hydrogen and oxygen. Then, multiply by the number of atoms of the element.

  • Hydrogen: 1.0079 g/mol
  • Oxygen: 15.9994 g/mol

There are 2 hydrogen atoms, so multiply the mass by 2.

  • 2 Hydrogen: (1.0079 g/mol)(2)= 2.0158 g/mol

Now, find the mass of H₂O. Add the mass of 2 hydrogen atoms and 1 oxygen atom.

  • 2.0158 g/mol + 15.9994 g/mol = 18.0152 g/mol

Next, find the amount of moles using the molecular mass we just calculated. Set up a ratio.

0.032 \ g  \ H_2 O* \frac{ 1 \ mol \ H_2 O}{18.0152 \ g \ H_2 O}

Multiply. The grams of H₂O will cancel out.

0.032 * \frac{1 \ mol \ H_2 O}{18.0152 }

\frac{0.032 *1 \ mol \ H_2 O}{18.0152 }

0.00177627781 \ mol \ H_2 O

The original measurement given had two significant figures (3,2). We must round to have 2 significant figures. All the zeroes before the 1 are not significant. So, round to the ten thousandth.

The 7 in the hundred thousandth place tells us to round up.

0.0018 \ mol \ H_2 O

There are about <u>0.0018 moles in 0.032 grams.</u>

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