Answer:
Ka = 4.76108
Explanation:
- CO(g) + 2H2(g) ↔ CH3OH(g)
∴ Keq = [CH3OH(g)] / [H2(g)]²[CO(g)]
[ ]initial change [ ]eq
CO(g) 0.27 M 0.27 - x 0.27 - x
H2(g) 0.49 M 0.49 - x 0.49 - x
CH3OH(g) 0 0 + x x = 0.11 M
replacing in Ka:
⇒ Ka = ( x ) / (0.49 - x)²(0.27 - x)
⇒ Ka = (0.11) / (0.49 - 0.11)² (0.27 - 0.11)
⇒ Ka = (0.11) / (0.38)²(0.16)
⇒ Ka = 4.76108
Answer is: Keq expression for this system is Keq = <span>[O</span>₂<span> ]</span> · [H₂<span>]</span>² ÷ [H₂O<span>]</span>².<span>
Chemical reaction: 2H</span>₂O(g) ⇄ O₂(g) + 2H₂(g).
The equilibrium constant<span> (Keq) is a ratio of the concentration of the products (in this reaction oxygen and hydrogen) to the concentration of the reactants (in this reaction water).</span>
One of the best buffer choice for pH = 8.0 is Tris with Ka value of 6.3 x 10^-9.
To support this answer, we first calculate for the pKa value as the negative logarithm of the Ka value:
pKa = -log Ka
For Tris, which is an abbreviation for 2-Amino-2-hydroxymethyl-propane-1,3 -diol and has a Ka value of 6.3 x 10^-9, the pKa is
pKa = -log Ka
= -log (6.3x10^-9)
= 8.2
We know that buffers work best when pH is equal to pKa:
pKa = 8.2 = pH
Therefore Tris would be a best buffer at pH = 8.0.
<span>This is due to the fact that the air pressure in that certain section of Earth’s atmosphere decreased. As density of gas particles decreases as air pressure decreases. Therefore, density of gas particles and air pressure have a direct relationship. An increase in air pressure would then effect to an increase in gas particles. </span>
