Answer:
11.72 grams
Explanation:
Let the equilibrium concentration of BrCl be y
Initial concentration of Br2 = number of moles ÷ volume = (0.979×1000/160) ÷ 199 = 0.031 M
Initial concentration of Cl2 = (1.075×1000/71) ÷ 199 = 0.076 M
From the equation of reaction
1 mole of Br2 reacted with 1 mole of Cl2 to form 2 moles of BrCl
Therefore, equilibrium concentration of Br2 = (0.031 - 0.5y) M while that of Cl2 = (0.076 - 0.5y) M
Kp = [BrCl]^2/[Br2][Cl2]
1.1×10^-4 = y^2/(0.031 - 0.5y)(0.076 - 0.5y)
y^2/0.002356-0.0535y+0.25y^2 = 0.00011
y^2/0.00011 = 0.002356-0.0536y+0.25y^2
9090.9y^2-0.25y^2+0.0536y-0.002356 = 0
9090.65y^2+0.0535y-0.002356 = 0
The value of y must be positive and is obtained using the quadratic formula
y = [-0.0535 + sqrt(0.0535^2 - 4×9090.65×-0.002356)] ÷ 2(9090.65) = 9.2025/18181.3 = 0.00051 M
Mass of BrCl = concentration×volume×MW = 0.00051×199×115.5 = 11.72 grams
First, we need to figure out the molar mass of P2O5.
P x 2: 30.97g x 2 = 61.94g
O x 5: 16.00g x 5 = 80.00g
+ 141.94g
0.450g x 1 mole/141.94g = 0.00317 moles = 3.17 x 10^3 moles
Now we are going to use Avogadro's number (=6.02 x 10^23)
(3.17 x 10^3 moles) x (6.02 x 10^23 atoms) / (1 mole) = 1.91 x 10^27 atoms
Answer:
Forward direction
Explanation:
The reaction quotient of an equilibrium reaction measures relative amounts of the products and the reactants present during the course of the reaction at particular point in the time.
Q < Kc , reaction will proceed in forward direction.
Q > Kc , reaction will proceed in backward direction.
Q = Kc , reaction at equilibrium.
It is the ratio of the concentration of the products and the reactants each raised to their stoichiometric coefficients. The concentration of the liquid and the gaseous species does not change and thus is not written in the expression.
Thus, for the reaction:
The expression is:
![Q=\frac {[CIO_3^{-}][Cl^{-}]^2}{[CIO^{-}]^3}](https://tex.z-dn.net/?f=Q%3D%5Cfrac%20%7B%5BCIO_3%5E%7B-%7D%5D%5BCl%5E%7B-%7D%5D%5E2%7D%7B%5BCIO%5E%7B-%7D%5D%5E3%7D)
Given,
[Cl⁻] = 0.50 mol/L; [ClO₃⁻] = 0.32 mol/L; [ClO⁻] = 0.24 mol/L
So,
Q = 5.7870
Since, Q < Kc (
)
The reaction will go in forward direction.
The value of E°cell and ∆G is 0.74V and 133.53kJ respectively.
node reaction (oxidation reaction)
Cd ------ Cd(+2) + 2e-
Cathode reaction (reduction reaction)
Cu(+2) + 2e- -------- Cu
Redox reaction
Cu(+2) + Cd ------- Cu + Cd(2+)
E°cell = E° cathode - E° anode
E°cathode = 0.34V
E°anode = -0. 40 V
E° cell = 0.34 - (-0.40)
E°cell = 0.74V
∆G° = - n FE° cell
= -2 × 96485 × 0.74
= -142798 J
= - 142.8 kJ
∆G° = -142. 8 kJ
∆G° = R T in K
= -142. 8 = - 8.314 × 10 ^ -3 × 298 × In k
Ink = 57.64
K = 1.08 × 10 ^25
b) Cd(s) + Cu+2 (aq) ------- Cd+2 (aq) + Cu (s)
E cell = E° cell - (0. 0591 /2) log (Cd+2 /Cu+2)
= 0.74V – (0.0591/2) log(1.95/0.05)
= 0.692 V.
∆G = -n F Ecell
= -2 × 96485 × 0.692
= 133.53kJ
Thus, we found that the value of E°cell and ∆G is 0.74V and 133.53kJ respectively.
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