Answer:
Kc = 0.075
Explanation:
The dissociation (α) is the initial quantity that ionized divided by the total dissolved. So, let's calling x the ionized quantity, and M the initial one:
α = x/M
x = M*α
x = 0.354M
For the stoichiometry of the reaction (2:1:1), the concentration of H₂ and I₂ must be half of the acid. So the equilibrium table must be:
2HI(g) ⇄ H₂(g) + I₂(g)
M 0 0 <em> Initial</em>
-0.354M +0.177M +0.177M <em>Reacts</em>
0.646M 0.177M 0.177M <em>Equilibrium</em>
The equilibrium constant Kc is the multiplication of the products' concentrations (elevated by their coefficients) divided by the multiplication of the reactants' concentrations (elevated by their coefficients):
![Kc = \frac{[H2]*[I2]}{[HI]^2}](https://tex.z-dn.net/?f=Kc%20%3D%20%5Cfrac%7B%5BH2%5D%2A%5BI2%5D%7D%7B%5BHI%5D%5E2%7D)
![Kc = \frac{0.177M*0.177M}{(0.646M)^2}](https://tex.z-dn.net/?f=Kc%20%3D%20%5Cfrac%7B0.177M%2A0.177M%7D%7B%280.646M%29%5E2%7D)
![Kc = \frac{0.03133M^2}{0.41732M^2}](https://tex.z-dn.net/?f=Kc%20%3D%20%5Cfrac%7B0.03133M%5E2%7D%7B0.41732M%5E2%7D)
Kc = 0.075
The formula for density is density =
![\frac{number of atoms in FCC * atomic weight}{ radius^{3}*Na}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bnumber%20of%20atoms%20in%20FCC%20%2A%20atomic%20weight%7D%7B%20radius%5E%7B3%7D%2ANa%7D%20)
. Substituting the given, the density is 162.69 g/cm3.
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Answer:- The pressure of ethanol would be 109 mmHg.
Solution:- This problem is based on Clausius clapeyron equation--
![ln(\frac{P_1}{P_2})=(\frac{\Delta Hvap}{R})(\frac{1}{T_2}-\frac{1}{T_1})](https://tex.z-dn.net/?f=ln%28%5Cfrac%7BP_1%7D%7BP_2%7D%29%3D%28%5Cfrac%7B%5CDelta%20Hvap%7D%7BR%7D%29%28%5Cfrac%7B1%7D%7BT_2%7D-%5Cfrac%7B1%7D%7BT_1%7D%29)
Given,
= 63.5 + 273 = 336.5 K
= 34.9 + 273 = 307.9 K
= 400 mmHg
= ?
= 39.3 kJ/mol = 39300 J/mol
R = 8.314 J/mol.K
Let's plug in the values in the equation and do the calculations.
![ln(\frac{400}{P_2})=(\frac{39300}{8.314})(\frac{1}{307.9}-\frac{1}{336.5})](https://tex.z-dn.net/?f=ln%28%5Cfrac%7B400%7D%7BP_2%7D%29%3D%28%5Cfrac%7B39300%7D%7B8.314%7D%29%28%5Cfrac%7B1%7D%7B307.9%7D-%5Cfrac%7B1%7D%7B336.5%7D%29)
= 1.30
On taking anti ln to both sides...
= ![e^1^.^3^0](https://tex.z-dn.net/?f=e%5E1%5E.%5E3%5E0)
= 3.67
= 400/3.67
= 109 mmHg