ΔG° at 450. K is -198.86kJ/mol
The following is the relationship between ΔG°, ΔH, and ΔS°:
ΔH-T ΔS = ΔG
where ΔG represents the common Gibbs free energy.
the enthalpy change, ΔH
The temperature in kelvin is T.
Entropy change is ΔS.
ΔG° = -206 kJ/mol
ΔH° equals -220 kJ/mol
T = 298 K
Using the formula, we obtain:
-220kJ/mol -T ΔS° = -206kJ/mol
220 kJ/mol +206 kJ/mol =T ΔS°.
-T ΔS = 14 kJ/mol
for ΔS-14/298
ΔS=0.047 kJ/mol.K
450K for the temperature Completing a formula with values
ΔG° = (450K)(-0.047kJ/mol)-220kJ/mol
ΔG° = -220 kJ/mol + 21.14 kJ/mol.
ΔG°=198.86 kJ/mol
Learn more about ΔG° here:
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Answer:
Kc = 3.1x10²
Explanation:
At equilibrium, the velocity of product formation is equal to the velocity of reactants formation. For a generic reaction, the equilibrium constant (Kc) is:
aA + bB ⇄ cC + dD
![Kc = \frac{[C]^c*[D]^d}{[A]^a*[B]^b}](https://tex.z-dn.net/?f=Kc%20%3D%20%5Cfrac%7B%5BC%5D%5Ec%2A%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%2A%5BB%5D%5Eb%7D)
Where [X] is the molar concentration of X, and the solid substances are not considered (because it's activity is 1, for the other substances, the activity is substituted for the molar concentration, which forms the equation above).
For the reaction given, let's make an equilibrium chart:
Fe³⁺(aq) + SCN⁻(aq) ⇄ FeSCN²⁺(aq)
1.1*10⁻³ 8.2*10⁻⁴ 0 <em> Initial</em>
-x -x +x <em>Reacts</em> (stoichiometry is 1:1:1)
1.1*10⁻³ -x 8.2*10⁻⁴ -x x <em> Equilibrium</em>
x = 1.8*10⁻⁴ M, so the molar concentrations at equilibrium are:
[Fe⁺³] = 1.1*10⁻³ - 1.8*10⁻⁴ = 9.2*10⁻⁴ M
[SCN⁻] = 8.2*10⁻⁴ - 1.8*10⁻⁴ = 6.4*10⁻⁴ M
[FeSCN⁺²] = 1.8*10⁻⁴ M
Kc = [FeSCN⁺²]/([Fe⁺³]*[SCN⁻])
Kc = (1.8*10⁻⁴)/(9.2*10⁻⁴*6.4*10⁻⁴)
Kc = 306 = 3.1x10²
The answer is A, light and heat energy.
Answer:
<h3>The answer is 7.47 g/cm³</h3>
Explanation:
The density of a substance can be found by using the formula

From the question
mass = 53.137 g
volume = 7.11 cm³
We have

We have the final answer as
<h3>7.47 g/cm³</h3>
Hope this helps you