What pressure, in atm, is exerted by 2.50 L of gas containing 1.35 mol at 320 K?
1 answer:
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Answer:
[C₆H₅COO⁻][H₃O⁺]/[C₆H₅COOH] = Ka
Explanation:
The reaction of dissociation of the benzoic acid in water is given by the following equation:
C₆H₅-COOH + H₂O ⇄ C₆H₅-COO⁻ + H₃O⁺ (1)
The dissociation constant of an acid is the measure of the strength of an acid:
HA ⇄ A⁻ + H⁺ (2)
(3)
<em>Where the dissociation constant of the acid (Ka) is equal to the ratio of the concentration of the dissociated forms of the acid, [A⁻][H⁺], and the concentration of the acid, [HA]. </em>
So, starting from the equations (2) and (3), the constant equation for the dissociation reaction of benzoic acid in water, of the equation (1), is:
I hope it helps you!
Use the formula for second order reaction:
C = concentration at time t
C0 = initial conc.
k = rate constant
t = time
1st equation :
2nd Equation:
Find
from 1st equation and put it in 2nd equation:
k = 0.046
C = concentration at time t
C0 = initial conc.
k = rate constant
t = time
1st equation :
2nd Equation:
Find
k = 0.046