Answer : The value of equilibrium constant for this reaction at 328.0 K is 
Explanation :
As we know that,

where,
= standard Gibbs free energy = ?
= standard enthalpy = 151.2 kJ = 151200 J
= standard entropy = 169.4 J/K
T = temperature of reaction = 328.0 K
Now put all the given values in the above formula, we get:


The relation between the equilibrium constant and standard Gibbs free energy is:

where,
= standard Gibbs free energy = 95636.8 J
R = gas constant = 8.314 J/K.mol
T = temperature = 328.0 K
K = equilibrium constant = ?
Now put all the given values in the above formula, we get:


Therefore, the value of equilibrium constant for this reaction at 328.0 K is 
All molecular motion stop at 0 k wich is zero kelvin. At absolute 0 it stops. The temperature of 0 entropy at which all molecular motion stops equals in centigrades to -273.15° C which is the same as 0 in kelvin degrees. Have in mind that t<span>emperature is a measure of the average kinetic energy of the </span>molecules<span> in a material.</span>
Answer:
A.) 4.0
Explanation:
The general equilibrium expression looks like this:
![K = \frac{[C]^{c} [D]^{d} }{[A]^{a} [B]^{b} }](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B%5BC%5D%5E%7Bc%7D%20%5BD%5D%5E%7Bd%7D%20%7D%7B%5BA%5D%5E%7Ba%7D%20%5BB%5D%5E%7Bb%7D%20%7D)
In this expression,
-----> K = equilibrium constant
-----> uppercase letters = molarity
-----> lowercase letters = balanced equation coefficients
In this case, the molarity's do not need to be raised to any numbers because the coefficients in the balanced equation are all 1. You can find the constant by plugging the given molarities into the equation and simplifying.
<----- Equilibrium expression
<----- Insert molarities
<----- Multiply
<----- Divide
Answer:
orbital
Explanation:
electrons are found in an orbital
The scientist's results is that at a temperature of 35<span>°C, the solubility of the substance in water is 146.2 grams in 200 grams of water. There isn't really a different method to determine the solubility of a substance in water. Another procedure could be that a lesser amount of the substance is used and the water required to dissolve it is determined. The solubility of the substance based on the two procedures can then be compared.</span>