Explanation:
The structure of Ferrarrisite Ca5(HAs O4)2(AsO4)2
Answer:
ΔG° = -5.4 kJ/mol
ΔG = 873.2 J/mol = 0.873 kJ /mol
Explanation:
Step 1: Data given
ΔG (NO2) = 51.84 kJ/mol
ΔG (N2O4) = 98.28 kJ/mol
Step 2:
ΔG = ΔG° + RT ln Q
⇒with Q = the reaction quatient
⇒with T = the temperature = 298 K
⇒with R = 8.314 J / mol*K
⇒with ΔG° = ΔG° (N2O4) - 2*ΔG°(NO2
)
⇒ ΔG° = 98.28 kJ/mol - 2* 51.84 kJ/mol
⇒ ΔG° = -5.4 kJ/mol
Part B
ΔG = ΔG° =RT ln Q
⇒with G° = -5.4 kj/mol = -5400 j/mol
⇒
with R = 8.314 J/K*mol
⇒with T = 298 K
⇒with Q = p(N2O4)/ [ p(NO2) ]² = 1.63/0.36² = 12.577
ΔG = -5400 + 8.314 * 298 * ln(12.577)
ΔG = -5400 + 8.314 * 298 * 2.532
ΔG = 873.2 J/mol = 0.873 kJ/mol
3 NaOH + FeBr3 = 3 NaBr + Fe(OH)3
Hope this helps!!
Answer:
hello, the answer is polygenic:)
Answer:
T = 215.33 °C
Explanation:
The activation energy is given by the Arrhenius equation:

<u>Where:</u>
k: is the rate constant
A: is the frequency factor
Ea: is the activation energy
R: is the gas constant = 8.314 J/(K*mol)
T: is the temperature
We have for the uncatalyzed reaction:
Ea₁ = 70 kJ/mol
And for the catalyzed reaction:
Ea₂ = 42 kJ/mol
T₂ = 20 °C = 293 K
The frequency factor A is constant and the initial concentrations are the same.
Since the rate of the uncatalyzed reaction (k₁) is equal to the rate of the catalyzed reaction (k₂), we have:

(1)
By solving equation (1) for T₁ we have:
Therefore, we need to heat the solution at 215.33 °C so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction.
I hope it helps you!