Answer:
9.2
Explanation:
Let's do an equilibrium chart of this reaction:
2NO(g) + O₂(g) ⇄ 2NO₂(g)
4.9 atm 5.1 atm 0 Initial
-2x -x +2x Reacts (stoichiometry is 2:1:2)
4.9-2x 5.1-x 2x Equilibrium
The mole fraction of NO₂ (y) can be calculated by the Raoult's law, that states that the mole fraction is the partial pressure divided by the total pressure:
y = 2x/(4.9 - 2x + 5.1 -x + 2x)
0.52 = 2x/(10 - x)
2x = 5.2 -0.52x
2.52x = 5.2
x = 2.06 atm
Thus, the partial pressure at equilibrium are:
pNO = 4.9 -2*2.06 = 0.78 atm
pO₂ = 5.1 - 2.06 = 3.04 atm
pNO₂ = 2*2.06 = 4.12 atm
Thus, the pressure equilibrium constant Kp is:
Kp = [(pNO₂)²]/[(pNO)²*(pO₂)]
Kp = [(4.12)²]/[(0.78)²*3.04]
Kp = [16.9744]/[1.849536]
Kp = 9.2
Answer:
Option E, Half life = 
Explanation:
For a first order reaction, rate constant and half-life is related as:
Where,
= Half life
k = Rate constant
Rate constant given = 
So, the correct option is option E.
Answer:
11.31g NaClO₂
Explanation:
<em> Is given 250mL of a 1.60M chlorous acid HClO2 solution. Ka is 1.110x10⁻². What mass of NaClO₂ should the student dissolve in the HClO2 solution to turn it into a buffer with pH =1.45? </em>
It is possible to answer this question using Henderson-Hasselbalch equation:
pH = pKa + log₁₀ [A⁻] / [HA]
<em>Where pKa is -log Ka = 1.9547; [A⁻] is the concentration of the conjugate base (NaClO₂), [HA] the concentration of the weak acid</em>
You can change the concentration of the substance if you write the moles of the substances:
[Moles HClO₂] = 250mL = 0.25L×(1.60mol /L) = <em>0.40 moles HClO₂</em>
Replacing in H-H expression, as the pH you want is 1.45:
1.45 = 1.9547 + log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
-0.5047 = log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
<em>0.3128 = </em>[Moles NaClO₂] / [0.40 moles HClO₂]
0.1251 = Moles NaClO₂
As molar mass of NaClO₂ is 90.44g/mol, mass of 0.1251 moles of NaClO₂ is:
0.1251 moles NaClO₂ ₓ (90.44g / mol) =
<h3>11.31g NaClO₂</h3>
