Answer:
(a) rate = -(1/3) Δ[O₂]/Δt = +(1/2) Δ[O₃]/Δt
(b) Δ[O₃]/Δt = 1.07x10⁻⁵ mol/Ls
Explanation:
By definition, t<u>he reaction rate for a chemical reaction can be expressed by the decrease in the concentration of reactants or the increase in the concentration of products:</u>
aX → bY (1)
![rate= -\frac{1}{a} \frac{\Delta[X]}{ \Delta t} = +\frac{1}{b} \frac{\Delta[Y]}{ \Delta t}](https://tex.z-dn.net/?f=%20rate%3D%20-%5Cfrac%7B1%7D%7Ba%7D%20%5Cfrac%7B%5CDelta%5BX%5D%7D%7B%20%5CDelta%20t%7D%20%3D%20%2B%5Cfrac%7B1%7D%7Bb%7D%20%5Cfrac%7B%5CDelta%5BY%5D%7D%7B%20%5CDelta%20t%7D%20)
<em>where, a and b are the coefficients of de reactant X and product Y, respectively. </em>
(a) Based on the definition above, we can express the rate of reaction (2) as follows:
3O₂(g) → 2O₃(g) (2)
(3)
(b) From the rate of disappearance of O₂ in equation (3), we can find the rate of appearance of O₃:
![rate = +\frac{1}{2} \frac{\Delta[O_{3}]}{ \Delta t} = -\frac{1}{3} \frac{\Delta[O_{2}]}{ \Delta t}](https://tex.z-dn.net/?f=%20rate%20%3D%20%2B%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7B%5CDelta%5BO_%7B3%7D%5D%7D%7B%20%5CDelta%20t%7D%20%3D%20-%5Cfrac%7B1%7D%7B3%7D%20%5Cfrac%7B%5CDelta%5BO_%7B2%7D%5D%7D%7B%20%5CDelta%20t%7D%20)
![\frac{\Delta[O_{3}]}{ \Delta t} = 1.07 \cdot 10^{-5} \frac{mol}{Ls}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5CDelta%5BO_%7B3%7D%5D%7D%7B%20%5CDelta%20t%7D%20%3D%201.07%20%5Ccdot%2010%5E%7B-5%7D%20%5Cfrac%7Bmol%7D%7BLs%7D%20)
So the rate of appearance of O₃ is 1.07x10⁻⁵ mol/Ls.
Have a nice day!
Answer:
7.28 × 10³ K
Explanation:
Let's consider the following reaction.
N₂(g) + O₂(g) → 2 NO(g)
The reaction is spontaneous when the standard Gibbs free energy (ΔG°) is negative. ΔG° is related to the standard enthalpy of the reaction (ΔH°) and the standard entropy of the reaction (ΔS°) through the following expression.
ΔG° = ΔH° - T . ΔS°
If ΔG° < 0,
ΔH° - T . ΔS° < 0
ΔH° < T . ΔS°
T > ΔH°/ΔS° = (180.5 × 10³ J/mol)/(24.8 J/mol.K) = 7.28 × 10³ K
The reaction is spontaneous above 7.28 × 10³ K.
We usually expess solubility as the maximum number of grams of of a solute that will dissolve in 100 g of solvent.
For example, at 20 °C, the solubility of <em>sodium chloride</em> is <em>35.89 g/100 g water</em>.
Answer:
d. 114.8 u
Explanation:
The formula for the calculation of the average atomic mass is:
Given that:
<u>For first isotope:
</u>
% = 95.72 %
Mass = 114.9041 u
<u>
For second isotope:
</u>
Since the element has only 2 isotopes, so the percentage of second is 100 - first percentage.
% = 100 % - 95.72 % = 4.28 %
Mass = 112.9043 u
Average Mass = ?
Thus,
<u>Average atomic mass = 114.8 u</u>