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Chapter 17: Problem 62

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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Most popular questions from this chapter

The equilibrium constant K for the reaction $$2 \mathrm{Cl}(g) \rightleftharpoons \mathrm{Cl}_{2}(g)$$ was measured as a function of temperature (Kelvin). A graph of \(\ln (K)\) versus 1\(/ T\) for this reaction gives a straight line with a slope of $1.352 \times 10^{4} \mathrm{K}\( and a y-intercept of \)-14.51 .$ Determine the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction. See Exercise \(85 .\)

Two crystalline forms of white phosphorus are known. Both forms contain \(\mathrm{P}_{4}\) molecules, but the molecules are packed together in different ways. The \(\alpha\) form is always obtained when the liquid freezes. However, below \(-76.9^{\circ} \mathrm{C},\) the \(\alpha\) form spontaneously converts to the \(\beta\) form: $$\mathrm{P}_{4}(s, \alpha) \longrightarrow \mathrm{P}_{4}(s, \beta)$$ a. Predict the signs of \(\Delta H\) and \(\Delta S\) for this process. b. Predict which form of phosphorus has the more ordered crystalline structure (has the smaller positional probability).

Consider the reaction: $$\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)$$ At \(25^{\circ} \mathrm{C}, \Delta G^{\circ}=-92.50 \mathrm{kJ}\) Which of the following statements is (are) true? a. This is an endothermic reaction. b. \(\Delta S^{\circ}\) for this reaction is negative. c. If the temperature is increased, the ratio \(\frac{\mathrm{PCl}_{5}}{\mathrm{PCl}_{3}}\) will increase. d. \(\Delta G^{\circ}\) for this reaction has to be negative at all temperatures. e. When \(\Delta G^{\circ}\) for this reaction is negative, then \(K\) is greater than 1.00 .

List three different ways to calculate the standard free energy change, \(\Delta G^{\circ},\) for a reaction at \(25^{\circ} \mathrm{C}\) . How is $\Delta G^{\circ}\( estimated at temperatures other than \)25^{\circ} \mathrm{C} ?$ What assumptions are made?

Which of the following reactions (or processes) are expected to have a negative value for \(\Delta S^{\circ} ?\) a. $\mathrm{SiF}_{6}(a q)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)+\mathrm{SiF}_{4}(g)$ b. $4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s)$ c. \(\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{COCl}_{2}(g)\) d. $\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)$ e. \(\mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\)
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