Answer:
The partial pressure of SO₃ is 82.0 atm
Explanation:
The equilibrium constant Kp is equal to <em>the equilibrium pressure of the gaseous products raised to the power of their stoichiometric coefficients divided by the equilibrium pressure of the gaseous reactants raised to the power of their stoichiometric coefficients</em>.
For the reaction,
2 SO₂(g) + O₂(g) → 2 SO₃(g)
![Kp = 0.345 = \frac{(pSO_{3})^{2} }{(pSO_{2})^{2} \times pO_{2} }\\pSO_{3} = \sqrt[]{0.345 \times (pSO_{2})^{2} \times pO_{2} } \\pSO_{3} = \sqrt[]{0.345 \times (35.0)^{2} \times 15.9 } \\pSO_{3} = 82.0 atm](https://tex.z-dn.net/?f=Kp%20%3D%200.345%20%3D%20%5Cfrac%7B%28pSO_%7B3%7D%29%5E%7B2%7D%20%7D%7B%28pSO_%7B2%7D%29%5E%7B2%7D%20%5Ctimes%20pO_%7B2%7D%20%7D%5C%5CpSO_%7B3%7D%20%3D%20%5Csqrt%5B%5D%7B0.345%20%5Ctimes%20%28pSO_%7B2%7D%29%5E%7B2%7D%20%5Ctimes%20pO_%7B2%7D%20%7D%20%5C%5CpSO_%7B3%7D%20%3D%20%5Csqrt%5B%5D%7B0.345%20%5Ctimes%20%2835.0%29%5E%7B2%7D%20%5Ctimes%2015.9%20%7D%20%5C%5CpSO_%7B3%7D%20%3D%2082.0%20atm)
Answer:
The new molar concentration of CO at equilibrium will be :[CO]=1.16 M.
Explanation:
Equilibrium concentration of all reactant and product:
![[CO_2] = 0.24 M, [H_2] = 0.24 M, [H_2O] = 0.48 M, [CO] = 0.48 M](https://tex.z-dn.net/?f=%5BCO_2%5D%20%3D%200.24%20M%2C%20%5BH_2%5D%20%3D%200.24%20M%2C%20%5BH_2O%5D%20%3D%200.48%20M%2C%20%5BCO%5D%20%3D%200.48%20M)
Equilibrium constant of the reaction :
![K=\frac{[H_2O][CO]}{[CO_2][H_2]}=\frac{0.48 M\times 0.48 M}{0.24 M\times 0.24 M}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BH_2O%5D%5BCO%5D%7D%7B%5BCO_2%5D%5BH_2%5D%7D%3D%5Cfrac%7B0.48%20M%5Ctimes%200.48%20M%7D%7B0.24%20M%5Ctimes%200.24%20M%7D)
K = 4
Concentration at eq'm:
0.24 M 0.24 M 0.48 M 0.48 M
After addition of 0.34 moles per liter of 
and 
are added.
(0.24+0.34) M (0.24+0.34) M (0.48+x)M (0.48+x)M
Equilibrium constant of the reaction after addition of more carbon dioxide and water:
Solving for x: x = 0.68
The new molar concentration of CO at equilibrium will be:
[CO]= (0.48+x)M = (0.48+0.68 )M = 1.16 M
Answer:
g NaCl = 424.623 g
Explanation:
<em>C</em> NaCl = 3.140 m = 3.140 mol NaCl / Kg solvent
∴ solvent: H2O
∴ mass H2O = 2.314 Kg
mol NaCl:
⇒ mol NaCl = (3.140 mol NaCl/Kg H2O)×(2.314 Kg H2O) = 7.266 mol NaCl
∴ mm NaCl = 58.44 g/mol
⇒ g NaCl = (7.266 mol NaCl)×(58.44 g/mol) = 424.623 g NaCl
Answer:
9.1
Explanation:
Step 1: Calculate the basic dissociation constant of propionate ion (Kb)
Sodium propionate is a strong electrolyte that dissociates according to the following equation.
NaC₃H₅O₂ ⇒ Na⁺ + C₃H₅O₂⁻
Propionate is the conjugate base of propionic acid according to the following equation.
C₃H₅O₂⁻ + H₂O ⇄ HC₃H₅O₂ + OH⁻
We can calculate Kb for propionate using the following expression.
Ka × Kb = Kw
Kb = Kw/Ka = 1.0 × 10⁻¹⁴/1.3 × 10⁻⁵ = 7.7 × 10⁻¹⁰
Step 2: Calculate the concentration of OH⁻
The concentration of the base (Cb) is 0.24 M. We can calculate [OH⁻] using the following expression.
[OH⁻] = √(Kb × Cb) = √(7.7 × 10⁻¹⁰ × 0.24) = 1.4 × 10⁻⁵ M
Step 3: Calculate the concentration of H⁺
We will use the following expression.
Kw = [H⁺] × [OH⁻]
[H⁺] = Kw/[OH⁻] = 1.0 × 10⁻¹⁴/1.4 × 10⁻⁵ = 7.1 × 10⁻¹⁰ M
Step 4: Calculate the pH of the solution
We will use the definition of pH.
pH = -log [H⁺] = -log 7.1 × 10⁻¹⁰ = 9.1
