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

User Phernost
by
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1 Answer

2 votes

Answer:

T = 377.2 K, Less than

Step-by-step 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.

User Akshay Arora
by
7.3k points
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