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?

Short Answer

Expert verified
\(\Delta G^{\circ} = 77.99 \, \mathrm{kJ} \, \mathrm{mol}^{-1}\)

Step by step solution

01

Convert the temperature from Celsius to Kelvin

The problem specifies a temperature of \(25^{\circ} \mathrm{C}\). To convert this to Kelvin, add 273.15, therefore, \( T = 25 + 273.15 = 298.15 \, \mathrm{K}\).
02

Substitute the given values into the equation

The equation for \(\Delta G^{\circ}\) is \(\Delta G^{\circ} = -RT \ln K\). Substituting \(\ln K_{\mathrm{w}} = \ln (1.0 \times 10^{-14})\), \(R = 8.314 \, \mathrm{J} \, \mathrm{mol}^{-1} \, \mathrm{K}^{-1}\) (as is standard in such calculations) and \(T = 298.15 \, \mathrm{K}\), the equation becomes \(\Delta G^{\circ} = -8.314 \times 298.15 \times \ln (1.0 \times 10^{-14})\).
03

Solve for \(\Delta G^{\circ}\)

Perform the multiplication and the natural logarithm operation to find the numerical value: \(\Delta G^{\circ} = -8.314 \times 298.15 \times (-31.42) = 77.99 \, \mathrm{kJ} \, \mathrm{mol}^{-1}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What is the role of ATP in biological reactions?

Explain the difference between \(\Delta G\) and \(\Delta G^{\circ}\).

Large quantities of hydrogen are needed for the synthesis of ammonia. One preparation of hydrogen involves the reaction between carbon monoxide and steam at \(300^{\circ} \mathrm{C}\) in the presence of a copper-zinc catalyst: $$ \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) $$.

The internal combustion engine of a \(1200-\mathrm{kg}\) car is designed to run on octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right),\) whose enthalpy of combustion is \(5510 \mathrm{~kJ} / \mathrm{mol}\). If the car is moving up a slope, calculate the maximum height (in meters) to which the car can be driven on 1.0 gallon of the fuel. Assume that the engine cylinder temperature is \(2200^{\circ} \mathrm{C}\) and the exit temperature is \(760^{\circ} \mathrm{C},\) and neglect all forms of friction. The mass of 1 gallon of fuel is \(3.1 \mathrm{~kg} .\) [Hint: The efficiency of the internal combustion engine, defined as work performed by the engine divided by the energy input, is given by \(\left(T_{2}-T_{1}\right) / T_{2},\) where \(T_{2}\) and \(T_{1}\) are the engine's operating temperature and exit temperature (in kelvins). The work done in moving the car over a vertical distance is \(m g h,\) where \(m\) is the mass of the car in \(\mathrm{kg}, g\) the acceleration due to gravity \(\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right),\) and \(h\) the height in meters. \(]\)

For each pair of substances listed here, choose the one having the larger standard entropy value at \(25^{\circ} \mathrm{C}\). The same molar amount is used in the comparison. Explain the basis for your choice. (a) \(\operatorname{Li}(s)\) or \(\operatorname{Li}(l)\) (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) or \(\mathrm{CH}_{3} \mathrm{OCH}_{3}(l)\) (c) \(\operatorname{Ar}(g)\) or \(\operatorname{Xe}(g)\) (d) \(\mathrm{CO}(g)\) or \(\mathrm{CO}_{2}(g)\) (e) \(\mathrm{O}_{2}(g)\) or \(\mathrm{O}_{3}(g)\) (f) \(\mathrm{NO}_{2}(g)\) or \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free