The value of ΔG° at this temperature is -18034.18 J/mol
Calculation,
Given information
formation constant (Kf)= 1.7 ×
Universal gas constant (R) = 8.314 J/K• mol
Temperature = 25° C = 25 °C + 273 = 300 K
Formula used:
ΔG° = -RT㏑Kf
By putting the valur of R,T, Kf we get the value of ΔG°
ΔG° = - 8.314 J/K• mol×300K㏑ 1.7 ×
ΔG° = -2494.2㏑ 1.7 × = -18034.18 J/mol
So, change in standard Gibbs's free energy is -18034.18 J/mol
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INFORMATION:
We have the following statements
And we must complete them
STEP BY STEP EXPLANATION:
To complete the statements, we need to classify matter according to its state:
- Solid:
there is not enough thermal energy to overcome the intermolecular interactions between the particles. As a result, solids have a definite shape and volume.
- Liquid:
That describes the liquid state. In a liquid, the particles are still in close contact, so liquids have a definite volume. However, because the particles can move about each other rather freely, a liquid has no definite shape and takes a shape dictated by its container.
- Gas:
That describes the gas state, which we will consider in more detail elsewhere. Like liquids, gases have no definite shape, but unlike solids and liquids, gases have no definite volume either.
Finally, we know that:
- A solid has a definite volume and has a definite shape
- A liquid has a definite volume and has not a definite shape
- A gas has not a definite volume and has not a definite shape
ANSWER:
- A solid has a definite volume and has a definite shape
- A liquid has a definite volume and has not a definite shape
- A gas has not a definite volume and has not a definite shape
