Answer:
ΔG°rxn = -69.0 kJ
Explanation:
Let's consider the following thermochemical equation.
N₂O(g) + NO₂(g) → 3 NO(g) ΔG°rxn = -23.0 kJ
Since ΔG°rxn < 0, this reaction is exergonic, that is, 23.0 kJ of energy are released. The Gibbs free energy is an extensive property, meaning that it depends on the amount of matter. Then, if we multiply the amount of matter by 3 (by multiplying the stoichiometric coefficients by 3), the ΔG°rxn will also be tripled.
3 N₂O(g) + 3 NO₂(g) → 9 NO(g) ΔG°rxn = -69.0 kJ
Answer:
The order of reaction is 2.
Rate constant is 0.0328 (M s)⁻¹
Explanation:
The rate of a reaction is inversely proportional to the time taken for the reaction.
As we are decreasing the concentration of the reactant the half life is increasing.
a) For zero order reaction: the half life is directly proportional to initial concentration of reactant
b) for first order reaction: the half life is independent of the initial concentration.
c) higher order reaction: The relation between half life and rate of reaction is:
Rate = ![\frac{1}{k[A_{0}]^{(n-1)}}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bk%5BA_%7B0%7D%5D%5E%7B%28n-1%29%7D%7D)
Half life =![K\frac{1}{[A_{0}]^{(n-1)} }](https://tex.z-dn.net/?f=K%5Cfrac%7B1%7D%7B%5BA_%7B0%7D%5D%5E%7B%28n-1%29%7D%20%7D)
![\frac{(halflife_{1})}{(halflife_{2})}=\frac{[A_{2}]^{(n-1)}}{[A_{1}]^{(n-1)} }](https://tex.z-dn.net/?f=%5Cfrac%7B%28halflife_%7B1%7D%29%7D%7B%28halflife_%7B2%7D%29%7D%3D%5Cfrac%7B%5BA_%7B2%7D%5D%5E%7B%28n-1%29%7D%7D%7B%5BA_%7B1%7D%5D%5E%7B%28n-1%29%7D%20%7D)
where n = order of reaction
Putting values
![\frac{109}{231}=\frac{[0.132]^{(n-1)}}{[0.280]^{(n-1)}}](https://tex.z-dn.net/?f=%5Cfrac%7B109%7D%7B231%7D%3D%5Cfrac%7B%5B0.132%5D%5E%7B%28n-1%29%7D%7D%7B%5B0.280%5D%5E%7B%28n-1%29%7D%7D)

Hence n = 2
![halflife=\frac{1}{k[A_{0}]}](https://tex.z-dn.net/?f=halflife%3D%5Cfrac%7B1%7D%7Bk%5BA_%7B0%7D%5D%7D)
Putting values

K = 0.0328
<u>Answer:</u> The volume of the container needed is 554.6 L
<u>Explanation:</u>
To calculate the volume of the gas, we use the equation given by ideal gas which follows:

where,
P = pressure of the gas = 3.0 atm
V = Volume of the gas = ? L
T = Temperature of the gas = ![25^oC=[25+273]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B25%2B273%5DK%3D298K)
R = Gas constant = 
n = number of moles of propane gas = 68.0 moles
Putting values in above equation, we get:

Hence, the volume of the container needed is 554.6 L
Answer: Cutting it, boiling it , freezing it
Explanation:
Mole ratio of Ethane to Oxygen = 2 : 7
Moles of O2 needed = 24 moles * (7/2) = 84 moles.