0
$\begingroup$

Case 1:

$$\ce{CH_3COOH + NaOH<=>CH_3COONa + H_2O}$$

At equilibrium, there remain extremely small concentrations of the reactants (acetic acid and sodium hydroxide), and comparatively large concentrations of the products (sodium acetate and water). So, what are the rates of reaction at equilibrium and at other times?

My thought process:

At equilibrium, the rates of the forward and the backward reactions become equal. So, we can see that compared to the products concentrations, very small concentrations of the reactants are enough for the forward reaction rate to equal the backward reaction rate, when large amount of products are present. So, when similar amount of products and reactants are present, the rate of the forward reaction will be much higher than the rate of the backward reaction. Intuitively,

$$\text{forward reaction rate(small concentrations)=backward reaction rate(large concentrations)[at equilibrium]}$$

$$\text{forward reaction rate(similar concentrations)>>backward reaction rate(similar concentrations)[not at equilibrium]}$$

I think I'm correct here.

Case 2:

$$\ce{NH_3 + H_2O<=>NH_4^{+} + OH^{-}}$$

At equilibrium, there remain large concentrations of the reactants (ammonia and water), and much smaller concentrations of the products (ammonium and hydroxyl ions). So, what are the rates of reaction at equilibrium and at other times?

My thought process:

At equilibrium, the rates of the forward and the backward reactions become equal. So, we can see that compared to the products concentrations, very large concentrations of the reactants are necessary for the forward reaction rate to equal the backward reaction rate, when much smaller amount of products are present. So, when similar amount of products and reactants are present, the rate of the forward reaction will be much lower than the rate of the backward reaction. Intuitively,

$$\text{forward reaction rate(large concentrations)=backward reaction rate(small concentrations)[at equilibrium]}$$

$$\text{forward reaction rate(similar concentrations)<<backward reaction rate(similar concentrations)[not at equilibrium]}$$

My problem:

In the comment section of this post, @Poutnik generously pointed out that my explanation of case 2 is wrong.

The stronger the base and lower pKb, the weaker the acid and higher pKa. And vice versa. Both pKa/b cannot be small. Ammonium dissociation NH4+ <=> NH3 + H+ ( pKa ) is much weaker then ammonia dissociation NH3+ H2O <=> NH4+ + OH- (pKb )

@Poutnik Kind sir, if the ammonium dissociation is weaker than the ammonia dissociation, then wouldn't that mean that the backward reaction rate is lesser than the forward reaction rate in the following reaction? NH3(aq)+H2O(l)↽−−⇀NH4+(aq)+OH−(aq) However, that's not the case. Here the forward reaction rate is lesser than the backward reaction rate. I'm a bit confused by your explanation kind sir. If you have time, could you please explain it to me?

It is the case. Ka = [H+][NH3]/[NH4+] = Kw.[NH3]/([OH-][NH4+]) = Kw/Kb.

I'm very confused now. All my life I thought that my explanation was correct. What do I do now?

$\endgroup$
6
  • 1
    $\begingroup$ I may somehow missed originally you talk about kinetics, not thermodynamics. // If you mean kinetic rates, then forward and backward reaction rates are equal at equilibrium ( reaction quotient is equal to equilibrium constant ). If reaction quotient is less than equilibrium constant, then the forward rate is greater than the backward rate. And vice versa. $\endgroup$
    – Poutnik
    Nov 26 '21 at 13:44
  • $\begingroup$ But saying the above, I add that reaction rates being equal at equlibrium say nothing about the position of the equilibrium itself, which is defined by the respective dissociation constants. $\endgroup$
    – Poutnik
    Nov 26 '21 at 13:50
  • $\begingroup$ @Poutnik In case 2, the reaction is shifted heavily to the left, right? $\endgroup$
    – user
    Nov 26 '21 at 14:03
  • 1
    $\begingroup$ Please make sure your title uniquely and concisely reflects your question, or is the question itself. A complaint with a request makes a poor and clickbait'y header. $\endgroup$
    – andselisk
    Nov 26 '21 at 14:17
  • 1
    $\begingroup$ The core of the issue seems to be confusion about reaction of ammonium with H2O vs with HO-, which is the same as saying the effect of pH on the rate of dissociation of ammonium. In your case 2, ammonia is reacting with water, so the back reaction is reaction of hydroxide with ammonium. But if the hydroxide concentration is very low (eg neutral or acidic pH), ammonium reacts primarily with water: $\ce{NH4+ + H2O <=> NH3 + H3O+}$ rather than $\ce{NH4+ + HO- <=> NH3 + H2O}$. The equilibrium of the first reaction favors the left side. That of the second favors the right. $\endgroup$
    – Andrew
    Nov 26 '21 at 15:21
2
$\begingroup$

Let say the kinetic rate of neutralization

$$\ce{CH3COOH + OH- -> CH3COO- + H2O}$$

is $k_\mathrm{f}[\ce{CH3COOH}][\ce{OH-}]$

And the kinetic rate of hydrolysis is

$$\ce{CH3COO- + H2O -> CH3COOH + OH-}$$

is $k_\mathrm{b}[\ce{CH3COO-}][\ce{H2O}]=k_\mathrm{b}^{*}[\ce{CH3COO-}]$

Then the equilibrium constant of the neutralization is $K = k_f/k_b^*$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.