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Question: The \({{K}_{a}}\) for \(C{{H}_{3}}COOH\) at \(300K\And 310K\) are \(1.8\times {{10}^{-5}}\) and \(1....

The Ka{{K}_{a}} for CH3COOHC{{H}_{3}}COOH at 300K&310K300K\And 310K are 1.8×10−51.8\times {{10}^{-5}} and 1.805×10−51.805\times {{10}^{-5}}, respectively. The enthalpy of deprotonation for acetic acid is 51.6cal⁡51.6\operatorname{cal}.
A. True
B. False

Explanation

Solution

The two values of the acid dissociation constant are for the same reaction but at different temperatures, which means they have the same entropy. Use the van’t Hoff equation here.

Complete answer:
First let’s be familiar with all the given conditions. There are two acid dissociation constants given; the only difference between them being the temperature at which they were recorded. And the enthalpy of acid dissociation of acetic acid is already given.
There is only one equation which relates the equilibrium constant (which in this case is the acid dissociation constant) to Gibbs free energy. It is as below:
ΔG=−RTln⁡Ka\Delta G=-RT\ln {{K}_{a}}
Where ΔG\Delta Grepresents the change in the free energy of a system, “R” represents the universal gas constant, “T” is the temperature and “K” is the equilibrium constant.
Now we also know an equation which relates the change in free energy (ΔG\Delta G) of a system to its change in enthalpy (ΔH\Delta H) and change in entropy (ΔS\Delta S) at a particular temperature (T). It is as below:
ΔG=ΔH−TΔS\Delta G=\Delta H-T\Delta S
Equating (1.1) and (1.2) we get:
ΔH−TΔS=−RTln⁡Ka\Delta H-T\Delta S=-RT\ln {{K}_{a}}
As there are two separate reactions at two different temperatures, we get the following taking this fact into account:
ΔH−T1ΔS=−RT1ln⁡Ka1\Delta H-{{T}_{1}}\Delta S=-R{{T}_{1}}\ln {{K}_{{{a}_{1}}}}
If we keep the equilibrium constant on the left hand side, the equation becomes:
ln⁡Ka1=ΔSR−ΔHRT1\ln {{K}_{{{a}_{1}}}}=\dfrac{\Delta S}{R}-\dfrac{\Delta H}{R{{T}_{1}}}
Similarly for the other equation, it becomes:
ln⁡Ka2=ΔSR−ΔHRT2\ln {{K}_{{{a}_{2}}}}=\dfrac{\Delta S}{R}-\dfrac{\Delta H}{R{{T}_{2}}}
Subtracting (1.5) from (1.6) we get:

& \ln \dfrac{{{K}_{{{a}_{2}}}}}{{{K}_{{{a}_{1}}}}}=\dfrac{\Delta S}{R}-\dfrac{\Delta H}{R{{T}_{2}}}-\dfrac{\Delta S}{R}+\dfrac{\Delta H}{R{{T}_{1}}} \\\ & \Rightarrow \ln \dfrac{{{K}_{{{a}_{2}}}}}{{{K}_{{{a}_{1}}}}}=\dfrac{\Delta H}{R}\left( \dfrac{1}{{{T}_{1}}}-\dfrac{1}{{{T}_{2}}} \right) \\\ \end{aligned}$$ The above equation which we derived is the van’t Hoff equation. We take reaction $1$ at $300\operatorname{K}$ and reaction $2$ at $310\operatorname{K}$. Putting the respective value in equation (1.7) we get: $$\ln \dfrac{1.805\times {{10}^{-5}}}{1.8\times {{10}^{-5}}}=\dfrac{\Delta H}{1.987}\left( \dfrac{1}{300}-\dfrac{1}{310} \right)$$ On solving equation (1.8) we get $16.038\operatorname{cal}$ which is not equal to the value given in the question. **Therefore the answer option (B) is false.** **Note:** It is important to remember that to derive the van’t Hoff equation we have to keep the equilibrium constant on the left hand side and all other variables to the right. If we do not do that we arrive at a different equation which is shown below: $$\ln \dfrac{{{K}_{2}}}{{{K}_{1}}}=\dfrac{RT}{\Delta S({{T}_{1}}-{{T}_{2}})}$$ Here we do not have the change in enthalpy value and therefore this form is not desirable if questions are related to the same.