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.
Answer:
2.2 °C/m
Explanation:
It seems the question is incomplete. However, this problem has been found in a web search, with values as follow:
" A certain substance X melts at a temperature of -9.9 °C. But if a 350 g sample of X is prepared with 31.8 g of urea (CH₄N₂O) dissolved in it, the sample is found to have a melting point of -13.2°C instead. Calculate the molal freezing point depression constant of X. Round your answer to 2 significant digits. "
So we use the formula for <em>freezing point depression</em>:
In this case, ΔTf = 13.2 - 9.9 = 3.3°C
m is the molality (moles solute/kg solvent)
- 350 g X ⇒ 350/1000 = 0.35 kg X
- 31.8 g Urea ÷ 60 g/mol = 0.53 mol Urea
Molality = 0.53 / 0.35 = 1.51 m
So now we have all the required data to <u>solve for Kf</u>:
First let us determine the electronic configuration of
Bromine (Br). This is written as:
Br = [Ar] 3d10 4s2 4p5
Then we must recall that the greatest effective nuclear
charge (also referred to as shielding) greatly increases as distance of the
orbital to the nucleus also increases. So therefore the electron in the
farthest shell will experience the greatest nuclear charge hence the answer is:
<span>4p orbital</span>
Answer:
8.70 liters
Explanation:
First we <u>convert 36.12 g of AI₂O₃ into moles</u>, using its <em>molar mass</em>:
- 36.12 g ÷ 101.96 g/mol = 0.354 mol AI₂O₃
Then we <u>convert AI₂O₃ moles into O₂ moles</u>, using the stoichiometric coefficients of the reaction:
- 0.354 mol AI₂O₃ *
= 0.531 mol O₂
We can now use the <em>PV=nRT equation</em> to <u>calculate the volume</u>, V:
- 1.4 atm * V = 0.531 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 280.0 K
Mass = number of mol x molar mass
Mass = 0.28mol x 55.8g/mol
Mass = 15.624g