Question
Use the equation in Exercise 79 to determine $\Delta H^{\circ}$ and $\Delta S^{\circ}$ for the autoionization of water: $$\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)$$
Step 1
This equation is in the form of a straight line equation $y = mx + c$, where $y = \ln K$, $m = -\frac{\Delta H^{\circ}}{R}$, $x = \frac{1}{T}$, and $c = \frac{\Delta S^{\circ}}{R}$. Show more…
Show all steps
Your feedback will help us improve your experience
Lottie Adams and 83 other Chemistry 101 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Use the equation in Exercise 85 to determine $\Delta H^{\circ}$ and $\Delta S^{\circ}$ for the autoionization of water: $$\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)$$
Use the equation in Exercise 67 to determine $\Delta H^{\circ}$ and $\Delta S^{\circ}$ for the autoionization of water: $$ \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ \begin{tabular}{cc} $T\left({ }^{\circ} \mathrm{C}\right)$ & $\boldsymbol{K}_{\mathrm{w}}$ \\ \hline 0 & $1.14 \times 10^{-15}$ \\ 25 & $1.00 \times 10^{-14}$ \\ 35 & $2.09 \times 10^{-14}$ \\ $40 .$ & $2.92 \times 10^{-14}$ \\ $50 .$ & $5.47 \times 10^{-14}$ \end{tabular}
For the autoionization of water at $25^{\circ} \mathrm{C},$ $\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)$ $K_{\mathrm{w}}$ is $1.0 \times 10^{-14} .$ What is $\Delta G^{\circ}$ for the process?
Transcript
600,000+
Students learning Chemistry with Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD