Explanation:
The given reaction is as follows.
Initial : 0.160 0.160 0
Change : -x -x 2x
Equilibrium: 0.160 - x 0.160 - x x
It is given that ![[H_{2}]](https://tex.z-dn.net/?f=%5BH_%7B2%7D%5D)
= [0.160 - x] = 0.036 M
and, ![[I_{2}]](https://tex.z-dn.net/?f=%5BI_%7B2%7D%5D)
= [0.160 - x] = 0.036 M
so, x = (0.160 - 0.036) M
= 0.124 M
As, [HI] = 2x.
So, [HI] = 
= 0.248 M
As it is known that expression for equilibrium constant is as follows.
![K_{eq} = \frac{[HI]^{2}}{[H_{2}][I_{2}]}](https://tex.z-dn.net/?f=K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BHI%5D%5E%7B2%7D%7D%7B%5BH_%7B2%7D%5D%5BI_%7B2%7D%5D%7D)
= 
= 47.46
Thus, we can conclude that the equilibrium constant, Kc, for the given reaction is 47.46.
The sum of the masses of the reactants is equal to the sum of the masses of the products.
Answer:
76.03 °C.
Explanation:
Equation:
C2H5OH(l) --> C2H5OH(g)
ΔHvaporization = ΔH(products) - ΔH (reactants)
= (-235.1 kJ/mol) - (-277.7 kK/mol)
= 42.6 kJ/mol.
ΔSvaporization = ΔS(products) - ΔS(reactants)
= 282.6 J/K.mol - 160.6 J/K.mol
= 122 J/K.mol
= 0.122 kJ/K.mol
Using gibbs free energy equation,
ΔG = ΔH - TΔS
ΔG = 0,
T = ΔH/ΔS
T = 42.6/0.122
= 349.18 K.
Coverting Kelvin to °C,
= 349.18 - 273.15
= 76.03 °C.
Answer:
E cell = +1.95 V
Explanation:
At Anode : Oxidation reaction takes place
At Cathode : Reduction reaction takes place
The reaction with lower value of reduction potential will undergo Oxidation
E = -1.18 V
This equation undergo oxidation reaction and become:
Anode(Oxidation-Half) :
E = +1.18 V
Cathode(Reduction-Half) :
E =+0.77 V
To balance the reaction multiply reduction-Half with 2.We get :
Note that E is intensive property , do not multiply E of oxidation-half with 2
Ecell = 0.77 -(-1.18)
E = +1.95 V
PH = 0.1289<span> for </span>1.50<span> M solution of weak acid with Ka value of </span><span>.73</span>
