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katrin [286]
3 years ago
13

The standard free energy of formation, ΔG∘f, of a substance is the free energy change for the formation of one mole of the subst

ance from the component elements in their standard states. These values are applicable at 25 ∘C and are found in thermodynamic tables.
The value of ΔG∘f for a substance gives a measure of the thermodynamic stability with respect to the component elements. Negative values for a formation reaction indicate thermodynamic stability of the product. In other words, the compound formed does not spontaneously decompose back into the component elements. Positive values for a formation reaction indicate thermodynamic instability of the product. Thus, the compound will spontaneously decompose, though the rate may be slow.

The sign of ΔG∘f can be used to predict the feasibility of synthesizing a substance from its component elements. The standard free energy change for a reaction, ΔG∘, is a state function and can be calculated from the standard free energies of formation as follows:

ΔG∘rxn=∑npΔG∘f(products)−∑nrΔG∘f(reactants)

where np and nr represent the stoichiometric coefficients in the balanced chemical equation for the reactants and products respectively.

Based on the standard free energies of formation, which of the following reactions represent a feasible way to synthesize the product?

A. N2(g)+H2(g)→N2H4(g); ΔG∘f=159.3 kJ/mol
B. 2Na(s)+O2(g)→Na2O2(s); ΔG∘f=−451.0 kJ/mol
C. 2C(s)+2H2(g)→C2H4(g); ΔG∘f=68.20 kJ/mol
D. 2SO(g)+O2(g)→2SO2(g); ΔG∘f=−600.4 kJ/mol
Chemistry
1 answer:
OLEGan [10]3 years ago
3 0

Answer:

B. 2 Na(s) + O₂(g) → Na₂O₂(s); ΔG∘f=−451.0 kJ/mol

D. 2 SO(g) + O₂(g) → 2 SO₂(g); ΔG°f=−600.4 kJ/mol

Explanation:

The spontaneity of a reaction  is given by the value of the standard Gibbs free energy of the reaction (ΔG°rxn). The more negative is the ΔG°rxn, the more spontaneous is a reaction.

The ΔG°rxn can be calculated using the following expression:

ΔG°rxn = ∑np × ΔG°f(products) − ∑nr × ΔG°f(reactants)

By definition, the standard Gibbs free energy of formation of simple substances in their most stable state is zero. That is why, in the reaction of formation of a compound ΔG°rxn = ΔG°f(product).

<em>Based on the standard free energies of formation, which of the following reactions represent a feasible way to synthesize the product? </em>

<em>     A. N₂(g) + H₂(g) → N₂H₄(g); ΔG°f=159.3 kJ/mol. </em>

<em>     </em>Not feasible. ΔG°rxn = ΔG°f(product) > 0.

    <em>B. 2 Na(s) + O₂(g) → Na₂O₂(s); ΔG°f=−451.0 kJ/mol</em>

    Feasible. ΔG°rxn = ΔG°f(product) < 0.

    <em>C. 2 C(s) + 2 H₂(g) → C₂H₄(g); ΔG°f=68.20 kJ/mol</em>

    Not feasible. ΔG°rxn = ΔG°f(product) > 0.

    <em>D. 2 SO(g) + O₂(g) → 2 SO₂(g); ΔG°f=−600.4 kJ/mol</em>

    Feasible. ΔG°rxn = ΔG°f(product) < 0.

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Answer:

Gaseous phase exert more pressure on container.

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It is the force exerted by the substance on its container.

Mathematical expression:

P = F/A

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A = area

We know that there are phases of matter solid liquid and gas.

Gases don't have definite volume and shape and take the shape and volume of container in which it present.  Molecule of gases randomly move everywhere and occupy all available space.  The molecules of gas randomly move collide with each other and also the wall of container their kinetic energy also increased because of this collision and pressure is produced

While in case of liquid molecules are packed and they can not move freely thus did not exert pressure like gaseous molecules.

In case of solids molecules are tightly packed and form more compact structure. They can not move thus did not exert pressure to the wall of container.

Properties of gases:

Molecule of gases randomly move everywhere and occupy all available space.

Gases don't have definite volume and shape and take the shape and volume of container in which it present.

Their densities are very low as compared to the liquid and solids.

Gas molecules are at long distance from each other therefore by applying pressure gases can be compressed.

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Liquid have definite volume but don,t have definite shape.

Their densities are high as compared to the gases but low as compared to the solids.

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Solids have definite volume and shape.

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6 0
3 years ago
Calculate the amount of heat that must be absorbed by 10.0 g of ice at –20°C to convert it to liquid water at 60.0°C. Given: spe
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Answer:

The amount of heat to absorb is 6,261 J

Explanation:

Calorimetry is in charge of measuring the amount of heat generated or lost in certain physical or chemical processes.

The total energy required is the sum of the energy to heat the ice from -20 ° C to ice of 0 ° C, melting the ice of 0 ° C in 0 ° C water and finally heating the water to 60 ° C.

So:

  • Heat required to raise the temperature of ice from -20 °C to 0 °C

Being the sensible heat of a body the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous), the expression is used:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial).

In this case, m= 10 g, specific heat of the ice= 2.1 \frac{J}{g*C} and ΔT=0 C - (-20 C)= 20 C

Replacing: Q= 10 g*2.1 \frac{J}{g*C} *20 C and solving: Q=420 J

  • Heat required to convert 0 °C ice to 0 °C water

The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:

Q= m* ΔHfusion

In this case, being 1 mol of water= 18 grams: Q= 10 g*6.0 \frac{kJ}{mol} *\frac{1 mol of water}{18 g}= 3.333 kJ= 3,333 J (being kJ=1,000 J)

  • Heat required to raise the temperature of water from 0 °C to 60 °C

In this case the expression used in the first step is used, but being: m= 10 g, specific heat of the water= 4.18 \frac{J}{g*C} and ΔT=60 C - (0 C)= 60 C

Replacing: Q= 10 g*4.18 \frac{J}{g*C} *60 C and solving: Q=2,508 J

Finally, Qtotal= 420 J + 3,333 J + 2,508 J

Qtotal= 6,261 J

<u><em> The amount of heat to absorb is 6,261 J</em></u>

<u><em></em></u>

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