When it comes to equilibrium reactions, it useful to do ICE analysis. ICE stands for Initial-Change-Equilibrium. You subtract the initial and change to determine the equilibrium amounts which is the basis for Kc. Kc is the equilibrium constant of concentration which is just the ratio of products to reactant.
Let's do the ICE analysis
2 NH₃ ⇄ N₂ + 3 H₂
I 0 1.3 1.65
C +2x -x -3x
-------------------------------------
E 0.1 ? ?
The variable x is the amount of moles of the substances that reacted. You apply the stoichiometric coefficients by multiplying it by x. Now, we can solve x by:
Equilibrium NH₃ = 0.1 = 0 + 2x
x = 0.05 mol
Therefore,
Equilibrium H₂ = 1.65 - 3(0.05) = 1.5 molEquilibrium N₂ = 1..3 - 0.05 = 1.25 mol
For the second part, I am confused with the given reaction because the stoichiometric coefficients do not balance which violates the law of conservation of mass. But you should remember that the Kc values might differ because of the stoichiometric coefficient. For a reaction: aA + bB ⇄ cC, the Kc for this is
![K_{C} = \frac{[ C^{c} ]}{[ A^{a} ][ B^{b} ]}](https://tex.z-dn.net/?f=%20K_%7BC%7D%20%3D%20%5Cfrac%7B%5B%20C%5E%7Bc%7D%20%5D%7D%7B%5B%20A%5E%7Ba%7D%20%5D%5B%20B%5E%7Bb%7D%20%5D%7D%20)
Hence, Kc could vary depending on the stoichiometric coefficients of the reaction.
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Answer:
The answer to your question is n = 5, l = 2, m can be -2, -1, 0, 1 or 2
Explanation:
Data
orbital = 5d
values of n, l, m
Process
1.- Determine the value of n
n is the coefficient of the orbital, in this problem n = 5
2.- Determine the value of l
l takes values depending in the sublevel of energy,
if the sublevel is s then l = 0
p l = 1
d l = 2
f l = 3
For this problem l = 2
3.- Determine the value of m
when l = 2, m takes values of -2, - 1, 0, 1 or 2