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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: ΔG = ΔH - TΔS. Given ΔH° = 62 kJ and ΔG° = 19 kJ, we first calculate ΔS at 25°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 = (ΔH - ΔG) / ΔS ≈ (62 kJ - 19 kJ) / (0.144 kJ/K) ≈ 298 K or 25°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: ΔG = ΔH - TΔS
02

Rearrange the equation to solve for T

First, we need to rearrange the equation to solve for the temperature (T): T = (ΔH - ΔG) / ΔS
03

Substitute the given ΔH° and ΔG° values

Now we can substitute the known values of ΔH° (62 kJ) and ΔG° (19 kJ) into the equation: T = (62 kJ - 19 kJ) / ΔS
04

Calculate ΔS°

Before we can solve for T, we need to find ΔS. We can do this using the given values of ΔH and ΔG: ΔS = (ΔH - ΔG) / T at 25°C (which is 298.15 K) ΔS = (62 kJ - 19 kJ) / 298.15 K ΔS ≈ 0.144 kJ/K
05

Calculate the temperature at which iodine sublimes

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

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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.

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Most popular questions from this chapter

Which of the following processes are spontaneous? a. Salt dissolves in \(\mathrm{H}_{2} \mathrm{O}\) . b. A clear solution becomes a uniform color after a few drops of dye are added. c. Iron rusts. d. You clean your bedroom.

Is \(\Delta S_{\text { surt }}\) favorable or unfavorable for exothermic reactions? Endothermic reactions? Explain.

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 \(25^{\circ} \mathrm{C}\)

Consider the following reaction at 298 K: $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ An equilibrium mixture contains \(\mathrm{O}_{2}(g)\) and \(\mathrm{SO}_{3}(g)\) at partial pressures of 0.50 \(\mathrm{atm}\) and 2.0 \(\mathrm{atm}\) , respectively. Using data from Appendix \(4,\) determine the equilibrium partial pressure of \(\mathrm{SO}_{2}\) in the mixture. Will this reaction be most favored at a high or a low temperature, assuming standard conditions?

Consider the reaction $$\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)$$ a. Use \(\Delta G_{f}^{\circ}\) values in Appendix 4 to calculate \(\Delta G^{\circ}\) for this reaction. b. Is this reaction spontaneous under standard conditions at 298 \(\mathrm{K} ?\) c. The value of \(\Delta H^{\circ}\) for this reaction is 100 . kJ. At what temperatures is this reaction spontaneous at standard conditions? Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

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