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GREYUIT [131]
2 years ago
11

What can be said about the spontaneity of this reaction? cdiamond→cgraphite

Chemistry
2 answers:
-Dominant- [34]2 years ago
8 0

The forward reaction of C (diamond) -> C (graphite) is spontaneous at 298 K

<h3>Further explanation </h3>

Gibbs free energy is the maximum possible work given by chemical reactions at constant pressure and temperature. Gibbs free energy can be used to determine the spontaneity of a reaction

If the Gibbs free energy value is <0 (negative) then the chemical reaction occurs spontaneously. If the change in free energy is zero, then the chemical reaction is at equilibrium, if it is> 0, the process is not spontaneous

Free energy of reaction (G) is the sum of its enthalpy (H) plus the product of the temperature and the entropy (S) of the system

Can be formulated: (at any temperature)

 \large{\boxed{\bold{\Delta G=\Delta H-T.\Delta S}}}

or at (25 Celsius / 298 K, 1 atm = standard)

ΔG ° reaction = ΔG ° f (products) - ΔG ° f (reactants)

Under standard conditions:

∆G ° = ∆H ° - T∆S °

The value of °H ° can be calculated from the change in enthalpy of standard formation:

∆H ° (reaction) = ∑H ° (product) - ∑ H ° (reagent)

The value of ΔS ° can be calculated from standard entropy data

∆S ° (reaction) = ∑S ° (product) - ∑ S ° (reagent)

If we look at the standard free Gibbs energy for the transformation from diamond to graphite at 298 K:

∆G ° rxn = -2.90 kJ

then a negative value indicates that reaction

C (diamond) -> C (graphite) occurs spontaneously (forward reaction)

 

<h3>Learn more   </h3>

Delta H solution  

brainly.com/question/10600048  

an exothermic reaction  

brainly.com/question/1831525  

as endothermic or exothermic  

brainly.com/question/11419458  

an exothermic dissolving process  

brainly.com/question/10541336  

anastassius [24]2 years ago
7 0
The gibbs free energy of the reaction of diamond to graphite is equal to -2.90 kJ/mol. The free energy is negative which means that the reaction is spontaneous. Therefore, the forward reaction is favored. Hope this helps. Have a nice day.
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In order to become stable, Potassium, K, will _______ and have a resulting charge of _____.
svlad2 [7]

Answer:

D

Explanation:

it doesn't matter how many protons k has all the answers have electron with the positive charge when is negative the question is trying to distract you D is the only one that would use the electrons sign correctly

4 0
2 years ago
Chlorine has two naturally stable isotopes: 35Cl (34.968853 amu) and 37Cl (36.965903 amu). The natural abundance of each isotope
allochka39001 [22]

Answer:

35Cl ⇒ 34.968853 amu  has an abundance of 30.05%

Explanation:

The molar mass of chlorine, (which is the average of all its naturally stable isotope masses), is 36.36575 amu.

There are 2 naturally stable isotopes, this means together they have an abundance of 100%

The isotopes are:

35Cl ⇒ 34.968853 amu  has an abundance of X %

37Cl ⇒ 36.965903 amu  has an abundance of Y %

X + Y = 100%   OR X = 100% - Y

36.36575 = 34.968853X + 36.965903Y  

36.36575 = 34.968853(1-Y) + 36.936.96590365903Y

36.36575 = 34.968853 -34.968853Y + 36.965903Y

1.396897 = 1.99705Y

Y = 0.699 = 69.95%

X = 100-69.9 = 30.05%

To control, we can plug in the following equation:

34.968853 * 0.3005 + 36.965903 * 0.6995 = 36.3658

This means

37Cl ⇒ 36.965903 amu  has an abundance of 69.95 %

35Cl ⇒ 34.968853 amu  has an abundance of 30.05%

7 0
3 years ago
when you put a book on a table the table pushes on the book is that newton's 1st,2nd,or 3rd law of motion ? 
ohaa [14]
The table pushes the book and the book pushes the table
It's 3rd law
8 0
2 years ago
Use standard enthalpies of formation to calculate Δ H ∘ rxn for each reaction. MISSED THIS? Read Section 7.9; Watch KCV 7.9, IWE
Eva8 [605]

Answer:

Standard Heat of Reaction 1 = -136.2 kJ/mol

Standard Heat of Reaction 2 = -41.166 kJ/mol

Standard Heat of Reaction 3 = -136.07 kJ/mol

Standard Heat of Reaction 4 = 279.448kJ/mol

Explanation:

C₂H₄ (g) + H₂ (g) → C₂H₆ (g)

CO (g) + H₂O (g) → H₂ (g) + CO₂ (g)

3NO₂ (g) + H₂O (l) → 2HNO₃ (aq) + NO (g)

Cr₂O₃ (s) + 3CO (g) → 2Cr (s) + 3CO₂ (g)

The required standard heat of formation for each of the reactants and product above, as obtained from literature is listed below.

C₂H₄ (g), 52.5 kJ/mol

H₂ (g), 0 kJ/mol

C₂H₆ (g), -83.7 kJ/mol

CO (g), -110.525 kJ/mol

H₂O (g), -241.818 kJ/mol

H₂ (g), 0 kJ/mol

CO₂ (g), -393.509 kJ/mol

NO₂ (g), 33.2 kJ/mol

H₂O (l), -285.8 kJ/mol

HNO₃ (aq), -206.28 kJ/mol

NO (g), 90.29 kJ/mol

Cr₂O₃ (s), -1128.4 kJ/mol

CO (g), -110.525 kJ/mol

Cr (s), 0 kJ/mol

CO₂ (g), -393.509 kJ/mol

Note that

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

C₂H₄ (g) + H₂ (g) → C₂H₆ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (1×-83.7) = -83.7 kJ/mol

ΔH∘(reactants) = (1×52.5) + (1×0) = 52.5 kJ/mol

ΔH∘(rxn) = -83.7 - 52.5 = -136.2 kJ/mol

CO (g) + H₂O (g) → H₂ (g) + CO₂ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (1×0) + (1×-393.509) = -393.509 kJ/mol

ΔH∘(reactants) = (1×-110.525) + (1×-241.818) = -352.343 kJ/mol

ΔH∘(rxn) = -393.509 - (-352.343) = -41.166 kJ/mol

3NO₂ (g) + H₂O (l) → 2HNO₃ (aq) + NO (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (2×-206.28) + (1×90.29) = -322.27 kJ/mol

ΔH∘(reactants) = (3×33.2) + (1×-285.8) = -186.2 kJ/mol

ΔH∘(rxn) = -322.27 - (-186.2) = -136.07 kJ/mol

Cr₂O₃ (s) + 3CO (g) → 2Cr (s) + 3CO₂ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (2×0) + (3×-393.509) = -1,180.527 kJ/mol

ΔH∘(reactants) = (1×-1128.4) + (3×-110.525) = -1,459.975 kJ/mol

ΔH∘(rxn) = -1,180.527 - (-1,459.975) = 279.448 kJ/mol

Hope this Helps!!!

4 0
3 years ago
Calculate the solubility of o2 in water at a partial pressure of o2 of 120 torr at 25 ̊c. the henry's law constant for o2 at 25
Vladimir79 [104]

Answer:

1) 2.054 x 10⁻⁴ mol/L.

2) Decreasing the temperature will increase the solubilty of O₂ gas in water.

Explanation:

1) The solubility of O₂ gas in water:

  • We cam calculate the solubility of O₂ in water using Henry's law: <em>Cgas = K P</em>,
  • where, Cgas is the solubility if gas,
  • K is henry's law constant (K for O₂ at 25 ̊C is 1.3 x 10⁻³ mol/l atm),
  • P is the partial pressure of O₂ (P = 120 torr / 760 = 0.158 atm).
  • Cgas = K P = (1.3 x 10⁻³ mol/l atm) (0.158 atm) = 2.054 x 10⁻⁴ mol/L.

2) The effect of decreasing temperature on the solubility O₂ gas in water:

  • Decreasing the temperature will increase the solubilty of O₂ gas in water.
  • When the temperature increases, the solubility of O₂ gas in water will decrease because the increase in T will increase the kinetic energy of gas particles and increase its motion that will break intermolecular bonds and escape from solution.
  • Decreasing the temperature will increase the solubility of O₂ gas in water will because the kinetic energy of gas particles will decrease and limit its motion that can not break the intermolecular bonds and increase the solubility of O₂ gas.


6 0
2 years ago
Read 2 more answers
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