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Chapter 17: Problem 39

For ammonia \(\left(\mathrm{NH}_{3}\right)\), the enthalpy of fusion is \(5.65 \mathrm{~kJ} / \mathrm{mol}\) and the entropy of fusion is \(28.9 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). a. Will \(\mathrm{NH}_{3}(s)\) spontaneously melt at \(200 . \mathrm{K}\) ? b. What is the approximate melting point of ammonia?

Short Answer

Expert verified
a. Yes, NH₃(s) will spontaneously melt at 200 K since the calculated ∆G is -130 J/mol, which is negative. b. The approximate melting point of ammonia is 195.5 K.

Step by step solution

01

Write down the Gibbs free energy equation

The equation for Gibbs free energy is given by: ∆G = ∆H - T∆S where ∆G is the change in Gibbs free energy, ∆H is the change in enthalpy, T is the temperature in Kelvin, and ∆S is the change in entropy.
02

Substitute given values into the equation

We are given the enthalpy of fusion (∆H) for ammonia as 5.65 kJ/mol and the entropy of fusion (∆S) as 28.9 J/K⋅mol. The temperature, T, is given as 200 K. Substitute these values into the Gibbs free energy equation: ∆G = (5.65 kJ/mol) - (200 K)(28.9 J/K⋅mol) Note that the units for ∆H and ∆S are different. Convert ∆H to J/mol: 5.65 kJ/mol × (1000 J/1 kJ) = 5650 J/mol Now, substitute the converted ∆H value into the equation: ∆G = (5650 J/mol) - (200 K)(28.9 J/K⋅mol)
03

Calculate ∆G

Now, calculate the value of ∆G: ∆G = 5650 J/mol - (200 K)(28.9 J/K⋅mol) ≈ 5650 J/mol - 5780 J/mol ≈ -130 J/mol Since the calculated ∆G is negative, the melting of NH₃(s) at 200 K is spontaneous. #b. Finding the approximate melting point#
04

Set ∆G equal to zero

At the melting point, the change in Gibbs free energy, ∆G, equals zero. We can use the same Gibbs free energy equation from part a and set ∆G to zero: 0 = ∆H - T∆S
05

Solve for the temperature, T

Now, rearrange the equation and solve for the temperature, T: T = ∆H / ∆S Substitute the given values for ∆H and ∆S: T = (5650 J/mol) / (28.9 J/K⋅mol) ≈ 195.5 K The approximate melting point of ammonia is 195.5 K.

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Most popular questions from this chapter

The equilibrium constant for a certain reaction increases by a factor of \(6.67\) when the temperature is increased from \(300.0 \mathrm{~K}\) to \(350.0 \mathrm{~K} .\) Calculate the standard change in enthalpy \(\left(\Delta H^{\circ}\right)\) for this reaction (assuming \(\Delta H^{\circ}\) is temperature-independent).

For the process \(\mathrm{A}(l) \longrightarrow \mathrm{A}(\mathrm{g})\), which direction is favored by changes in energy probability? Positional probability? Explain your answers. If you wanted to favor the process as written, would you raise or lower the temperature of the system? Explain.

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