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Question: If \(C{{H}_{3}}COOH\) \(({{K}_{a}}={{10}^{-5}})\) reacts with \(NaOH\) at 298 K, then find out the v...

If CH3COOHC{{H}_{3}}COOH (Ka=10−5)({{K}_{a}}={{10}^{-5}}) reacts with NaOHNaOH at 298 K, then find out the value of the maximum rate constant of the reverse reaction at 298K at the end point of the reaction. Given that the rate constant of the forward reaction is 10−11 mol−1Lsec⁡−1{{10}^{-11}}\,mo{{l}^{-1}}L{{\sec }^{-1}}at 298K. also, calculate Arrhenius parameter for backward reaction, if ΔH298=44kcal and Ea(f)=94kcal\Delta {{H}_{298}}=44kcal\,\text{and}\,{{E}_{a(f)}}=94kcal :
(A) Kb=10−20,Ab=2.71×1016{{K}_{b}}={{10}^{-20}},{{A}_{b}}=2.71\times {{10}^{16}}
(B) Kb=10−19,Ab=5.4×1016{{K}_{b}}={{10}^{-19}},{{A}_{b}}=5.4\times {{10}^{16}}
(C) Kb=10−10,Ab=5.4×1014{{K}_{b}}={{10}^{-10}},{{A}_{b}}=5.4\times {{10}^{14}}
(D) None of these

Explanation

Solution

The reaction being a weak acid and strong base reaction, from the hydrolysis constant of the salt hydrolysis and the dissociation constant of the weak acid, the relation to the rate constant obtained and thus, the Arrhenius parameter is calculated.

Complete step by step solution:
It is given that a reaction between a weak acid, acetic acid (CH3COOH)(C{{H}_{3}}COOH) and strong base, sodium hydroxide (NaOH)(NaOH) takes place. It is a neutralization reaction, leading to the formation of sodium acetate salt and water, with the equilibrium constant Kc{{K}_{c}} , as follows:
CH3COOH+NaOH⇌CH3COONa+H2OC{{H}_{3}}COOH+NaOH\rightleftharpoons C{{H}_{3}}COONa+{{H}_{2}}O
Here, the forward rate constant is k1{{k}_{1}} and backward rate constant is k2{{k}_{2}}. So, the Kc{{K}_{c}} is given by:
Kc=k1k2{{K}_{c}}=\dfrac{{{k}_{1}}}{{{k}_{2}}} ------ (1)
The reverse of the above reaction is known as the hydrolysis reaction of the salt, which completely dissociates into its ions in the solution. The solution is basic, due to the presence of hydroxide ions, as the acetate (conjugate base) abstracts a proton from the water molecule, with a hydrolysis constant Kh{{K}_{h}} , the reaction is as follows:
CH3COONa→CH3COO−+Na+(spectator cation)C{{H}_{3}}COONa\to C{{H}_{3}}CO{{O}^{-}}+N{{a}^{+}}\text{(spectator}\,cat\text{ion)}
Hydrolysis: CH3COO−+H2O⇄CH3COOH+OH−C{{H}_{3}}CO{{O}^{-}}+{{H}_{2}}O\rightleftarrows C{{H}_{3}}COOH+O{{H}^{-}}
The hydrolysis constant of the reaction is,
Kh=  [CH3COOH][OH−][CH3COO−][H2O]   {{K}_{h}}=\,\,\dfrac{\left[ C{{H}_{3}}COOH \right]\left[ O{{H}^{-}} \right]}{\left[ C{{H}_{3}}CO{{O}^{-}} \right]\left[ {{H}_{2}}O \right]}\,\,\, ------- (2)
Then, as the equilibrium constant and the hydrolysis constant are inversely proportional. It is given as:
Kc=(Kh)−1=  [CH3COO−][H2O][CH3COOH][OH−]  {{K}_{c}}={{\left( {{K}_{h}} \right)}^{-1}}=\,\,\dfrac{\left[ C{{H}_{3}}CO{{O}^{-}} \right]\left[ {{H}_{2}}O \right]}{\left[ C{{H}_{3}}COOH \right]\left[ O{{H}^{-}} \right]}\,\, ------- (3)
It is also given that, the acetic acid with a Ka{{K}_{a}} , being a weak acid, it dissociates partially. We get the dissociation constant as,
Ka=  [CH3COO−][H+][CH3COOH]  =10−5{{K}_{a}}=\,\,\dfrac{\left[ C{{H}_{3}}CO{{O}^{-}} \right]\left[ {{H}^{+}} \right]}{\left[ C{{H}_{3}}COOH \right]}\,\,={{10}^{-5}}

We can get, [CH3COO−][CH3COOH]=10−5[H+]\dfrac{\left[ C{{H}_{3}}CO{{O}^{-}} \right]}{\left[ C{{H}_{3}}COOH \right]}=\dfrac{{{10}^{-5}}}{\left[ {{H}^{+}} \right]} ------- (4)
Also, as we know, the ionic product constant of water is:
KW=[H+][OH−][H2O]=10−14{{K}_{W}}=\dfrac{\left[ {{H}^{+}} \right]\left[ O{{H}^{-}} \right]}{\left[ {{H}_{2}}O \right]}={{10}^{-14}}
We can get, [H+][10−14]=[H2O][OH−]\dfrac{\left[ {{H}^{+}} \right]}{\left[ {{10}^{-14}} \right]}=\dfrac{\left[ {{H}_{2}}O \right]}{\left[ O{{H}^{-}} \right]} -------- (5)
Now, we substitute the values of equation (4) and (5) in equation (3), we get,
Kc=(Kh)−1= 10−5[H+] ×[H+]10−14  {{K}_{c}}={{\left( {{K}_{h}} \right)}^{-1}}=\,\dfrac{{{10}^{-5}}}{\left[ {{H}^{+}} \right]}\,\times \dfrac{\left[ {{H}^{+}} \right]}{{{10}^{-14}}}\,\,
Kc=k1k2= 10−5[H+] ×[H+]10−14  {{K}_{c}}=\dfrac{{{k}_{1}}}{{{k}_{2}}}=\,\dfrac{{{10}^{-5}}}{\left[ {{H}^{+}} \right]}\,\times \dfrac{\left[ {{H}^{+}} \right]}{{{10}^{-14}}}\,\,
Therefore, k2=10−20{{k}_{2}}={{10}^{-20}} (as given forward rate constant k1=10−11 mol−1Lsec⁡−1{{k}_{1}}={{10}^{-11}}\,mo{{l}^{-1}}L{{\sec }^{-1}}) --- (6)
Now, in order to find the Arrhenius parameter, Ab{{A}_{b}} for backward reaction given that, ΔH298=44kcal and Ea(f)=94kcal\Delta {{H}_{298}}=44kcal\,\text{and}\,{{E}_{a(f)}}=94kcal --------- (7)
As, the enthalpy ΔH= Ef−Eb\Delta H=\,{{E}_{_{f}}}-{{E}_{b}}
Substituting the values of (7) in this, we get, 44 kcal= 94 kcal−Eb44\,kcal=\,94\,kcal-{{E}_{b}}, that is,
Eb=50 kcal=50×1000×4.187 Joules{{E}_{b}}=50\,kcal=50\times 1000\times 4.187\,\text{Joules} --------- (8)
Then, from the Arrhenius equation, for the backward reaction,
k2=Abe−Eb/RT{{k}_{2}}={{A}_{b}}{{e}^{-{{E}_{b}}/RT}}
Taking log on both sides:
log⁡k2=log⁡Ab−EbRT\log {{k}_{2}}=\log {{A}_{b}}-\dfrac{{{E}_{b}}}{RT}
By substituting the values of equation (6) and (8) in the above equation, we get,
log⁡10−20=log⁡Ab−50×1000×4.1878.314×298\log {{10}^{-20}}=\log {{A}_{b}}-\dfrac{50\times 1000\times 4.187}{8.314\times 298}
Therefore, the Arrhenius parameter isAb=2.71×1016 L mol-1sec-1{{A}_{b}}=2.71\times {{10}^{16}}\,\text{L mo}{{\text{l}}^{\text{-1}}}\text{se}{{\text{c}}^{\text{-1}}} .

Thus, the backward rate constant and the Arrhenius parameter is given by option (A)- Kb=10−20,Ab=2.71×1016{{K}_{b}}={{10}^{-20}},{{A}_{b}}=2.71\times {{10}^{16}}.

Note: The hydrolysis constant and the dissociation constant of the acid is taken to be the ratio of the concentration of the product to the reactant, to the power of the coefficient which is the moles of the substance in the reaction.