Question

A) Use the equation in exercise 88 to determine $\mathit{\Delta }{\mathit{H}}^{\mathbf{o}}$and $\mathit{\Delta }{\mathit{S}}^{\mathbf{o}}$for the autoionization of water:

${\mathit{H}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{\left(}\mathbf{l}\mathbf{\right)}}\mathbf{\rightleftarrows }{{\mathit{H}}^{\mathbf{+}}}_{\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}}\mathbf{+}\mathit{O}{{\mathit{H}}^{\mathbf{-}}}_{\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}}$

$\frac{\mathbf{T}{\mathbf{(}}^{\mathbf{o}}\mathbf{C}\mathbf{)}\mathbf{K}}{\begin{array}{l}\mathbf{0}\mathbf{1}\mathbf{.}\mathbf{14}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{15}}\\ \mathbf{25}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{14}}\\ \mathbf{35}\mathbf{2}\mathbf{.}\mathbf{09}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{14}}\\ \mathbf{40}\mathbf{.}\mathbf{2}\mathbf{.}\mathbf{92}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{14}}\\ \mathbf{50}\mathbf{5}\mathbf{.}\mathbf{47}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{14}}\end{array}}$

B) Estimate the value of for the autoionization of water at its critical temperature,

Short answer

(a)The and is 57 J/Kand -77 J/Ksolution respectively

(b) 106.8 kJ is the value for

Step-by-step solution

Description for autoionization

On the given reaction,

$H_2{O}_{\left(l\right)}\rightleftarrows {H^{+}}_{\left(aq\right)}+O{H^{-}}_{\left(aq\right)}$

$T{(}^{\circ }C)$	$T{(}^{\circ }C)+273$	K	In$K$	$\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$T$}\right.$
0	$273$	$1.14\times {10}^{-15}$	$-34.41$	$0.003663004$
25	$298$	$1.00\times {10}^{-14}$	$-32.24$	$0.003355705$
35	$308$	$2.09\times {10}^{-14}$	$-31.50$	$0.003246753$
40	$313$	$2.92\times {10}^{-14}$	$-31.16$	$0.003194888$
50	$323$	$5.47\times {10}^{-14}$	$-30.54$	$0.003095975$

Determine ΔHoandΔSo

The equation obtained is$-6856.3x-9.2372$

In Slope,

$\begin{array}{l}\frac{-\Delta H^{\circ }}{R}=-6856.3\\ \Delta H^{\circ }=-6856.3\times 0.008314\\ \Delta H^{\circ }=57.0J/K\end{array}$

In Intercept,

$\begin{array}{l}\frac{\Delta S^{\circ }}{R}=-9.2372\\ \Delta S^{\circ }=-9.2372\times 0.008314\\ \Delta S^{\circ }=-77.0J/K\end{array}$

Determine ΔG0

$\begin{array}{l}\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }\\ =57kJ-(374+273)K(-0.077kJ/K)\\ =106.8kJ\end{array}$

Therefore, the ΔH⁰and ΔS⁰is 57 J/Kand -77 J/Ksolution respectively and also, 106.8 kJis the value for ΔG⁰