Water can be an atom.<br><br> True<br> False

Diamond is forever" is one of the most successful advertising slogans of all time. but is it true? for the reaction shown below,

Zielflug [23.3K]

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.

7 0

4 years ago

If a scientist accidentally spills a chemical that causes invalid conclusions, he has made__________________?

iren [92.7K]

It is B) An honest mistake

5 0

3 years ago

Read 2 more answers

Which of the following required Bohr's model of the atom to need modification ? A. Energies of electrons are quantized. B. Quant

ki77a [65]

Answer:

Electrons do not follow circular orbits around the nucleus

Explanation:

Bohr's model of the atom is a combination of elements of quantum theory and classical physics in approaching the problem of the hydrogen atom. According to Neils Bohr, stationary states exist in which the energy of the electron is constant. These stationary states were referred to as circular orbits which encompasses the nucleus of the atom. Each orbit is characterized by a principal quantum number (n). Energy is absorbed or emitted when an electron transits between stationary states in the atom.

Sommerfeld improved on Bohr's proposal by postulating that instead of considering the electron in circular orbits, electrons actually orbited around the nucleus in elliptical orbits, this became a significant improvement on Bohr's model of the atom until the wave mechanical model of Erwin Schrödinger was proposed.

7 0

3 years ago

A student produces 1.34g of mgnh4po4*6h2o (mm =245.1 g/mol) from a sample of fertilizer, how grams of phosphorous were in the fe

In-s [12.5K]

1.34 grams of MgNH4PO4*6H2O
Molar mass= 245.1 grams per mole
Moles of MgNH4PO4*6H2O=
(1.34 grams)(1 mole/245.1 grams fertilizer)=0.0054671563 moles fertilizer
MgNH4PO4*6H2O has 1 mole of Phosphorous, so:
0.0054671563 moles fertilizer(1 mole Phosphorous/1 mole fertilizer)
0.0054671563 moles Phosphorous (30.97 grams/ 1 mole)
=0.1693178295 grams Phosphorous

4 0

3 years ago

Read 2 more answers

How common is tryptophan in our food, what effect does it have on our body?

dedylja [7]

I can't give you an exact number or anything for how common it is. However, I can say that it is QUITE common in our food.

There's many effects it has to the human body, therefore I'll only list a few.

~social behavior~irritability~mood swings~sleepiness
Also, if tryptophan is used as a drug it can be used to treat depression. Though, it can only help to a certain extent. It varies as well; depending on the person. (Don't quote me on that completely though.)

I hope this helped :)

5 0

3 years ago

Water can be an atom. True False