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fredd [130]
3 years ago
9

A reaction is non-spontaneous at any temperature when

Chemistry
1 answer:
slavikrds [6]3 years ago
8 0

Answer:

A reaction is non-spontaneous at any temperature when the Gibbs free energy > 0.

Explanation:

There is a state function, that determines if a reaction is sponaneous or non spontaneous:

ΔG = Gibbs free energy

A reaction is non spontaneous when it does require energy to produce that reaction. It will be spontaneous, when the reaction does not require energy to be occured.

The formula is: ΔG = ΔH - T.ΔS

ΔH → Enthalpy → Energy gained or realeased as heat.

ΔH < 0 → <em>Exothermic reaction. Spontaneity is favored </em>

T  → Temperature

ΔS → Entropy →  Degree of disorder of a system.

When the system has a considered disorder ΔS > 0, disorder increases.

When the system is more ordered, ΔS < 0, disorder decreases.

The reaction will be non spontaneous if, the enthalpy is positive (endothermic reaction) and the ΔS < 0 (disorder decreases). It will not occur if we do not give energy.

ΔG < 0 → Spontaneous reaction

ΔG > 0 → Non spontaneous reaction

ΔG = 0 → System in equilibrium

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1) The densities of air at −85°C, 0°C, and 100°C are 1.877 g dm−3, 1.294 g dm−3, and 0.946 g dm−3, respectively. From these data
earnstyle [38]

Answer:

1) The absolute zero temperature is -272.74 °C

2) The absolute zero temperature is -269.91°C

Explanation:

1) The specific volume of air at the given temperatures are;

At -85°C, v =  1/(1.877) = 0.533 dm³/g, the pressure =

At 0°C, v = 1/1.294 ≈ 0.773 dm³/g

At 100°C, v = 1/0.946 ≈ 1.057 dm³/g

We therefore have;

v₁ = 0.533 T₁ = 188.15  K

v₂ = 0.773 T₂ = 273.15  K

v₃ = 1.057 T₃ = 373.15  K

The slope of the graph formed by the above data is therefore given as follows;

m = (1.057 - 0.533)/(373.15 - 188.5) = 0.00284 dm³/K

The equation is therefor;

v - 0.533 = 0.00284×(T - 188.15)

v = 0.00284×T - 0.534 + 0.533  = 0.00284×T - 0.001167

v =  0.00284×T - 0.001167

Therefore, when the temperature, at absolute 0, we have;

v = 0

Which gives;

0 =  0.00284×T - 0.001167

0.00284×T = 0.001167

T = 0.001167/0.00284 ≈ 0.411 K

Which is 0.411  -273.15 ≈ -272.74 °C

The absolute zero temperature is -272.74 °C

2) The given volume of the gas = 20.00 dm³ at 0°C and 1.000 atm

The slope of the volume temperature graph at constant pressure = 0.0741 dm³/(°C)

The temperature is converted to Kelvin temperature, in order to apply Charles law as follows;

0°C = 0 + 273.15 K = 273.15 K

Therefore, the equation of the graph can be presented as follows;

v - 20.00 = 0.0741 × (T - 273.15)

Which gives;

v = 0.0741·T - 0.0741 ×(273.15) + 20

v = 0.0741·T - 20.2404125 + 20

v = 0.0741·T - 0.240415

Therefore at absolute 0, v = 0, we have;

0 = 0.0741·T - 0.240415

0.0741·T = 0.240415

T = 0.240415/0.0741 = 3.2445 K

The temperature in degrees is therefore;

3.2445 K - 273.15 ≈ -269.91°C

The absolute zero temperature is therefore, -269.91°C

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