Answer:
1.9 x 10⁻¹⁴
Explanation:
The Gibbs-Helmholtz equation which gives us the dependendence of the equilibrium constant, k, at the different temperatures, T₁ and T₂, will be used to solve this question:
ln(k₂/k₁) = - ΔHº/R ( 1/T₂ - 1/T₁ )
Here we are to assume ΔHº is constant over the temperature range 25-100 ºC.
We have all the data input required, so lets substitute and solve for k₂
T₁ = (25 + 273) K = 298 K
T₂ = (53 +273) K = 326 K
R = 8.314 J
ln( k₂/1.0 x 10⁻¹⁴ ) = - 5.58 x 10⁴ J/K x (8.314 J/K ( 1/326 K - 1/298 K ))
ln( k₂/1.0 x 10⁻¹⁴ ) = 1.9
Notice the units cancel each other as we would expect it to be since k₂/k₁ is unitless.
Now take inverse ln to both sides of the equation:
k₂/ 1.0 x 10⁻¹⁴ = 1.93 ⇒ k₂ = 1.9 x 10⁻¹⁴
This result is logical because the reaction is endothermic given ΔH° is positive at 25 ºC, so at 53 ºC we would expect k₂ greater than k₁.
The autoionization of water kw at 53°C is 6.905 × 10⁻¹⁴
Autoionization of water refers to the transference of one water molecule to another for the purpose of producing a hydroxide ion as well as a hydronium ion.
At equilibrium, the autoionization of water can be expressed as:
where;
From the parameters given:
The first process is to calculate the free energy change associated with the autoionization of water;
[tex]\mathbf{\Delta G = -RTIn K_w} \\ \\ \mathbf{\Delta G = -8.314 J/K/mol\times 298.15 K \times In (1.0\times 10^{-14}) } \\ \\ \mathbf{\Delta G \simeq 79908 \ J/mol}[/tex]
Recall that:
Then, we can determine the entropy change for the process using the above parameters as:
[tex]\mathbf{\Delta G = \Delta H^0 - T\Delta S^0}[/tex]
[tex]\mathbf{79908 \ J/mol = 5.58 \times 10^4 \ J/mol - 298.15 \ K \times \Delta S^0}[/tex]
[tex]\mathbf{\Delta S^0 = \dfrac{79908 \ J/mol -5.58 \times 10^4 \ J/mol }{ - 298.15 \ K}}[/tex]
[tex]\mathbf{\Delta S^0 = -80.86 \ J/K/mol}[/tex]
However, if we made an assumption that [tex]\mathbf{\Delta S^0 }[/tex] is also constant, Then:
The Temperature at 53 °C = (53 + 273.15)K = 326.15 K
[tex]\mathbf{\Delta G = \Delta H^0 - T\Delta S^0}[/tex]
[tex]\mathbf{\Delta G = 5.58 \times 10^4 \ J/mol - 326.15 \ K \times-80.86 \ J/K/mol} \\ \\ \mathbf{\Delta G =82172.489 \ J/mol}[/tex]
Therefore, calculating the kw at 53°C, we have:
[tex]\mathbf{\Delta G = -RT In k_w}[/tex]
[tex]\mathbf{k_w = e^{-\dfrac{\Delta G}{RT}}}[/tex]
[tex]\mathbf{k_w = e^{-\dfrac{82172.489 \ J.mol }{8.314 \ J/K/mol \times 326.15 \ K}}}[/tex]
[tex]\mathbf{k_w = 6.905\times 10^{-14} }[/tex]
Therefore, we can conclude that the kw at 53°C is 6.905 × 10⁻¹⁴
Learn more about the autoionization of water here:
https://brainly.com/question/7251420
