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3 years ago

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

Chemistry

2 answers:

-Dominant- [34]3 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]3 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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fgiga [73]

This problem is providing a chemical reaction reaction at equilibrium and its pressure-based equilibrium constant, and thus asks for molecules of SO2 at equilibrium. At the end, the answer turns out to be 1.25x10¹² molecules.

<h3>Chemical equilibrium</h3>

In chemistry, when chemical reactions take place, they do not necessarily go to completion, but can reach a constant-concentration point denominated "equilibrium" where the reaction does not proceed any longer.

In such a way, for the given reaction, one can write the equilibrium expression based on the law of mass action, which sets the products on top, reactants on bottom and ignores both solids and liquids:

$Kp=\frac{1}{p_{SO_2}*p_{O_2}^{1/2}}$

However, in order to calculate this, one needs the partial pressure of O2, defined in similar problems as 0.21 atm for us to find the partial pressure of SO2 first:

$2.38x10^{15}=\frac{1}{p_{SO_2}*(0.21)^{1/2}}\\\\p_{SO_2}=\frac{1}{(2.38x10^{15})(0.21)^{1/2}}\\\\p_{SO_2}=9.17x10^{-16}atm$

Keep in mind this 0.21 atm may vary depending on the problem you were given.

Thus, we obtain the mole fraction of SO2 as follows:

$x_{SO_2}=\frac{9.17x10^{-16}atm}{9.17x10^{-16}atm+0.21atm}=4.37x10^{-15}\\ \\$

Finally, we use this mole fraction to calculate the moles of SO2 in 475 moles of the gas mixture, and then use the Avogadro's number to calculate the molecules:

$molecules=475mol*4.37x10^{-15}*\frac{6.022x10^{23}}{1mol} =1.25x10^{12}molecules$

Learn more about chemical equilibrium: brainly.com/question/26453983

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We can calculate the new volume of the gas using the Combined Gas Law:

(P1 x V1) / T1 = (P2 x V2) / T2

The initial volume, pressure, and temperature were 280 mL, 1.3 atm, and 291.15 K (changing the temperature into Kelvin is necessary), and the final volume, pressure, and temperature is V2, 3.0 atm, and 308.15 K. Plugging these values in and solving, we find that:

(P1 x V1) / T1 = (P2 x V2) / T2

(1.3 atm x 280 mL) / 291.15 K = (3.0 atm x V2) / 308.15 K

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This makes sense considering the conditions, a small increase in temperature would make the gas expand but a significant increase in the pressure would cause the volume to decrease.

Hope this helps!

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