\(\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{2} \mathrm{O}(s)\) Which of the following is true for the above reaction? (A) The value for \(\Delta S\) is positive. (B) The value for \(\Delta G\) is zero. (C) The value for \(\Delta H\) is positive. (D) The reaction is favored at 1.0 \(\mathrm{atm}\) and 298 \(\mathrm{K}\) .

Short Answer

Expert verified
Based on the information provided, only statement (A) is incorrect. Due to the decrease in disorder during the transformation from a liquid to a solid state, the entropy change \(\Delta S\) for this reaction is negative, not positive.

Step by step solution

01

Understanding Entropy

Entropy \(\Delta S\) represents the degree of randomness or disorder in a system. When a substance changes from liquid to solid its disorder decreases. Thus, for the given reaction \(\Delta S\) is negative.
02

Understanding Enthalpy

Enthalpy \(\Delta H\) represents the heat content of the system. For a substance to change from liquid to solid, it would release heat. Therefore, \(\Delta H\) is negative
03

Understanding Gibbs Free Energy

The Gibbs Free Energy \(\Delta G\) indicates whether a reaction is spontaneous or not. \(\Delta G\) is calculated as \(\Delta G = \Delta H - T\Delta S\), where \(T\) is the temperature. However, without specific temperature or numerical values for \(\Delta H\) and \(\Delta S\) in the problem, it's hard to definitively say if the \(\Delta G\) is zero or not.
04

Reaction Favorability at Given Conditions

Whether a reaction is favored under certain conditions (such as 1.0 atm and 298 K in this case) is best determined by the Gibbs Free Energy. However, as noted in the previous step, without specific values, we cannot clearly state. On a general note, substances tend to be in the solid state at 1.0 atm and 298 K.

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

\(\begin{array}{ll}{\mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=-390 \mathrm{kJ} / \mathrm{mol}} \\\ {\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-290 \mathrm{kJ} / \mathrm{mol}} \\ {2 \mathrm{C}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)} & {\Delta H^{\circ}=+230 \mathrm{kJ} / \mathrm{mol}}\end{array}\) Based on the information given above, what is \(\Delta H^{\circ}\) for the following reaction? $$ \begin{aligned} \mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \\ \text { (A) }-1,300 \mathrm{kJ} \\ \text { (B) }-1,070 \mathrm{kJ} \\ \text { (C) }-840 \mathrm{kJ} \\ \text { (D) }-780 \mathrm{kJ} \end{aligned} $$

Questions 32-36 refer to the following. Two half-cells are set up as follows: Half-Cell A: Strip of \(\mathrm{Cu}(s)\) in \(\mathrm{CuNO}_{3}(a q)\) Half-Cell B: Strip of \(\mathrm{Zn}(s)\) in \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) (aq) When the cells are connected according to the diagram below, the following reaction occurs: GRAPH CAN'T COPY $$2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) E^{\circ}=+1.28 \mathrm{V}$$ As the reaction progresses, what will happen to the overall voltage of the cell? (A) It will increase as \(\left[\mathrm{Zn}^{2+}\right]\) increases. (B) It will increase as \(\left[\mathrm{Cu}^{+}\right]\) increases. (C) It will decrease as \(\left[\mathrm{Zn}^{2+}\right]\) increases. (D) The voltage will remain constant.

If the solubility of \(\mathrm{BaF}_{2}\), is equal to \(x,\) which of the following expressions is equal to the solubility product, \(K_{s p}\) , for \(\operatorname{BaF}_{2} ?\) (A) \(x^{2}\) (B) 2\(x^{2}\) (C) 2\(x^{3}\) (D) 4\(x^{3}\)

The following reaction is found to be at equilibrium at 25°C: \(2 \mathrm{SO}_{3}(g) \leftrightarrow \mathrm{O}_{2}(g)+2 \mathrm{SO}_{2}(g) \quad \Delta H=-198 \mathrm{kJ} / \mathrm{mol}\) Which of the following would cause a reduction in the value for the equilibrium constant? (A) Increasing the amount of \(\mathrm{SO}_{3}\) (B) Reducing the amount of \(\mathrm{O}_{2}\) (C) Raising the temperature (D) Lowering the temperature

After 44 minutes, a sample of \(_{19}^{44} \mathrm{K}\) is found to have decayed to 25 percent of the original amount present. What is the half-life of \(_{19}^{44} \mathrm{K}?\) (A) 11 minutes (B) 22 minutes (C) 44 minutes (D) 66 minutes

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