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For a particular reaction, ΔH∘=−28.4 kJ and ΔS∘=−87.9 J/K. Assuming these values change very little with temperature, at what temperature does the reaction change from nonspontaneous to spontaneous?

Respuesta :

Answer:

323 K

Explanation:

Since it is in equilibrium, we get  Δ=0=Δ−Δ

                                                      Δ=Δ

                                                      =ΔΔ

Now we solve for T

T = ( -28.4 kJ) / (-87.9 J/K * 1kJ/1000J) = 323 K

Parrain

Based on the fact that there is very little change in value with temperature, the temperature that the reaction changes to spontaneous would be 323.1K.

What temperature causes a reaction change to be spontaneous?

This can be found as:

ΔH∘ = Temperature x  ΔS∘

Temperture = ΔH∘ /  ΔS∘

Solving gives:

= - 28.4 / - 87.9

= 323.1 k

Find out more on spontaneous reactions at https://brainly.com/question/13700386.

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