/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 62 For the sublimation of iodine at... [FREE SOLUTION] | 91影视

91影视

For the sublimation of iodine at \(25^{\circ} \mathrm{C}\) $$\mathrm{I}_{2}(s) \rightarrow \mathrm{I}_{2}(g)$$ the values of \(\Delta H^{\circ}\) and \(\Delta G^{\circ}\) are, respectively, 62 \(\mathrm{kJ}\) and 19 \(\mathrm{kJ}\) . Estimate the temperature at which iodine sublimes. Assume \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

Short Answer

Expert verified
The temperature at which iodine sublimes can be determined using the Gibbs-Helmholtz equation: 螖G = 螖H - T螖S. Given 螖H掳 = 62 kJ and 螖G掳 = 19 kJ, we first calculate 螖S at 25掳C (298.15 K) to be approximately 0.144 kJ/K. Then, we can find the sublimation temperature T by substituting the given values into the rearranged equation: T = (螖H - 螖G) / 螖S 鈮 (62 kJ - 19 kJ) / (0.144 kJ/K) 鈮 298 K or 25掳C.

Step by step solution

01

Write the Gibbs-Helmholtz equation

The formula we will be using to determine the temperature at which iodine sublimes is the Gibbs-Helmholtz equation: 螖G = 螖H - T螖S
02

Rearrange the equation to solve for T

First, we need to rearrange the equation to solve for the temperature (T): T = (螖H - 螖G) / 螖S
03

Substitute the given 螖H掳 and 螖G掳 values

Now we can substitute the known values of 螖H掳 (62 kJ) and 螖G掳 (19 kJ) into the equation: T = (62 kJ - 19 kJ) / 螖S
04

Calculate 螖S掳

Before we can solve for T, we need to find 螖S. We can do this using the given values of 螖H and 螖G: 螖S = (螖H - 螖G) / T at 25掳C (which is 298.15 K) 螖S = (62 kJ - 19 kJ) / 298.15 K 螖S 鈮 0.144 kJ/K
05

Calculate the temperature at which iodine sublimes

Now that we have calculated 螖S, we can substitute that value into the equation for T: T = (62 kJ - 19 kJ) / (0.144 kJ/K) T 鈮 298 K The temperature at which iodine sublimes is approximately 298 K or 25掳C.

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 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sublimation of Iodine
Iodine is an interesting element that undergoes sublimation rather than melting. This means that instead of going from solid to liquid, it changes directly from a solid state to a gas. Sublimation is a fascinating physical process often observed in elements like iodine and dry ice (solid carbon dioxide). This characteristic property makes iodine useful in various scientific experiments and practical applications. When you heat iodine, it skips the liquid stage and forms a violet-to-blue gas. This process happens under normal atmospheric pressure and needs relatively low amounts of energy compared to melting. Understanding sublimation involves looking at energy and molecular interactions that can be well explained through thermodynamics, including concepts like enthalpy and entropy changes.
Enthalpy Change
Enthalpy change, denoted as \( \Delta H \), is a crucial concept when discussing chemical processes. It represents the total heat content or energy change in a system during a chemical reaction or phase change at constant pressure. In the case of iodine sublimation, the enthalpy change is the energy required to convert solid iodine into iodine gas. The value provided, \( \Delta H^{\circ} = 62 \) kJ, indicates that iodine requires this much energy per mole to sublime at a standard temperature. Key points about enthalpy change:
  • Positive \( \Delta H \): Energy is absorbed, typical of endothermic processes like sublimation.
  • The larger the \( \Delta H \), the more energy is needed for the phase change.
Understanding \( \Delta H \) helps predict how much energy will be used or released during a reaction. For iodine, the positive value means that sublimation requires energy input, crucial for estimating conditions at which it occurs.
Gibbs Free Energy
Gibbs free energy, \( \Delta G \), is a thermodynamic quantity that predicts the spontaneity of a process. It combines the concepts of enthalpy, entropy, and temperature to give a full picture of energy changes. The formula is given by: \( \Delta G = \Delta H - T\Delta S \)For iodine sublimation at \( 25^{\circ} \text{C} \), \( \Delta G^{\circ} = 19 \) kJ indicates how much free energy is available for the phase transition. A positive \( \Delta G \) suggests the process is non-spontaneous under given conditions, needing energy input to proceed.Key insights into Gibbs free energy include:
  • \( \Delta G < 0 \): Reaction is spontaneous.
  • \( \Delta G = 0 \): System is at equilibrium.
  • \( \Delta G > 0 \): Reaction needs energy, as seen in sublimation.
This concept helps to predict phase transitions and the conditions needed for these to occur.
Entropy Change
Entropy change, \( \Delta S \), measures the change in disorder or randomness in a system during a process. It's an essential part of the Gibbs-Helmholtz equation, which combines to determine the spontaneity of reactions. In sublimation, where a solid changes to a gas, \( \Delta S \) is usually positive. This means the randomness of iodine molecules increases as they spread out from a solid to a gaseous state. Calculated from the given values, \( \Delta S^{\circ} \approx 0.144 \text{ kJ/K} \), indicates an increase in disorder as iodine transitions. Highlights about entropy change:
  • Positive \( \Delta S \): System becomes more disordered.
  • Negative \( \Delta S \): System becomes more ordered.
  • Influences whether a process is spontaneous along with \( \Delta H \) and temperature \( T \).
By evaluating \( \Delta S \), we can understand molecular behavior changes, crucial to thermodynamic processes like sublimation.

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

You remember that \(\Delta G^{\circ}\) is related to \(R T \ln (K)\) but cannot remember if it's \(R T \ln (K)\) or \(-R T \ln (K)\) . Realizing what \(\Delta G^{\circ}\) and K mean, how can you figure out the correct sign?

The third law of thermodynamics states that the entropy of a perfect crystal at 0 \(\mathrm{K}\) is zero. In Appendix \(4, \mathrm{F}^{-}(a q), \mathrm{OH}^{-}(a q)\) and \(\mathrm{S}^{2-}(a q)\) all have negative standard entropy values. How can \(S^{\circ}\) values be less than zero?

Consider the reaction: $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightarrow 2 \mathrm{NO}_{2}(g)$$ where \(P_{\mathrm{NO}_{2}}=0.29\) atm and \(P_{\mathrm{N}_{2} \mathrm{O}_{4}}=1.6 \mathrm{atm} .\) For this reaction at these conditions, \(\Delta G=-1000 \mathrm{J}\) and \(\Delta G^{\circ}=6000 \mathrm{J}\) . Which of the following statements about this reaction is(are) true? a. The reverse reaction is spontaneous at these conditions. b. At equilibrium, \(P_{\mathrm{N}_{2} \mathrm{O}_{4}}\) will be greater than 1.6 \(\mathrm{atm}\) . c. The value of K for this reaction is greater than 1. d. The maximum amount of work this reaction can produce at these conditions is 鈥6000 J. e. The reaction is endothermic.

Ethanethiol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{SH} ; \text { also called ethyl mercaptan) is }\right.\) commonly added to natural gas to provide the 鈥渞otten egg鈥 smell of a gas leak. The boiling point of ethanethiol is \(35^{\circ} \mathrm{C}\) and its heat of vaporization is 27.5 \(\mathrm{kJ} / \mathrm{mol}\) . What is the entropy of vaporization for this substance?

Consider the dissociation of a weak acid \(\mathrm{HA}\left(K_{\mathrm{a}}=4.5 \times 10^{-3}\right)\) in water: $$\mathrm{HA}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{A}^{-}(a q)$$ Calculate \(\Delta G^{\circ}\) for this reaction at \(25^{\circ} \mathrm{C}\)

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.