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storchak [24]
3 years ago
13

For a particular reaction, ΔH⁰ = − 46.7 kJ/mol and ΔS⁰ = − 123.8 J / (mol ⋅ K). Assuming these values change very little with te

mperature, at what temperature does the reaction change from non-spontaneous to spontaneous in the forward direction?
T= ______K
Is the reaction in the forward direction spontaneous at temperatures greater than or less than the calculated temperature?
a. less than
b. greater than
Chemistry
1 answer:
PilotLPTM [1.2K]3 years ago
8 0

Answer:

T = 377.2 K, Less than

Explanation:

The thermodynamic quantity used in predicting whether a reaction is spontaneous or not is the gibbs free energy.

It's relationship with ΔH⁰ and ΔS⁰ is given as;

ΔG° = ΔH° - TΔS°

Basically, a negative value of ΔG° means the reaction is spontaeneous.

To obtain the calculated vale of T,

ΔS° = ΔH°/T

T = ΔH° / ΔS°

T = 377.2 K

Let's calculate the value of ΔG° at that temperature.

ΔG° = ΔH° - TΔS°

ΔG° =  − 46700 - 377.2(− 123.8)

ΔG°  = 0 (approximately, values are due to the rounding off)

At ΔG°  = 0 the reaction is at equilibrium.

To find if the reaction is spontaneous at lower or hugher temperature than the calculated temperature, we would be substituting the value of T with a smaller (random) value and also a larger (random) value.

Larger T (390K)

ΔG° = ΔH° - TΔS°

ΔG° =  − 46700 - 390(− 123.8)

ΔG°  = - 46700 + 48,282

ΔG° = 1582 J/mol

Smaller T (350K)

ΔG° = ΔH° - TΔS°

ΔG° =  − 46700 - 350(− 123.8)

ΔG°  = - 46700 + 43330

ΔG° = -3370J/mol

This means the temperature would be lesser than the calculated value for it to be spontaneus.

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Use the fact that to determine how much the pressure must change in order to lower the boiling point of water by a small amount
erma4kov [3.2K]

Complete Question

Use the fact that d\mu=\frac{V}{N}dp-\frac{s}{N}dT to determine how much the pressure must change in order to lower the boiling point of water by a small amount 3.20e-01 K. You may assume that the entropy and density of the liquid and gas are roughly constant for these small changes. You may also assume that the volume per molecule of liquid water is approximately zero compared to that of water vapor, and that water vapor is an ideal gas. Useful constants: Atmospheric pressure is 101300 Pa The boiling point of water at atmospheric pressure is 373.15 K The entropy difference between liquid and gas per kilogram is 6.05e 03 J/kgK The molecular weight of water is 0.018 kg/mol. (a) 0.00e 00 Pa (b) 1.14e 03 Pa (c) 6.85e 26 Pa (d) 4.24e 05 Pa (e) 3.81e 28 Pa

Answer:

Correct option is B

Explanation:

From the question we are told that:

Given Equation d\mu=\frac{V}{N}dp-\frac{s}{N}dT

Change of boiling point \triangle H=3.20e-01 K

Generally the equation for Change in time is mathematically given by

  d\mu=\frac{V}{N}dp-\frac{s}{N}dT

  dp=\frac{s}{v}dT

Where

    s=Entropy\ difference *molar\ weight

    s=6.05*10^3*0.018j/mol.k

And

    V=\frac{RT}{P} (from ideal gas equation)

Therefore

 dp=\frac{Ps}{RT}dT

 dp=\frac{101300*6.05*10^3*0.018}{8.314*373.15}3.20*10^{-1}

 dp=1137.873pa

 dp=1.14e 03 Pa

Therefore correct option is B

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