A reaction will be spontaneous only at low temperatures if both ΔH and ΔS are negative. For a reaction in which ΔH = −310.1 kJ/mol and ΔS = −89.00 J/K · mol, determine the temperature (in °C) below which the reaction is spontaneous.

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Respuesta :

Mergus Mergus · Answer 1 · 2019-11-26T07:29:29+01:00

Answer:

3211.12 K

Explanation:

The expression for the standard change in free energy is:

[tex]\Delta G=\Delta H-T\times \Delta S[/tex]

Where,

[tex]\Delta G[/tex] is the change in the Gibbs free energy.

T is the absolute temperature. (T in kelvins)

[tex]\Delta H[/tex] is the enthalpy change of the reaction.

[tex]\Delta S[/tex] is the change in entropy.

For reaction to be spontaneous, [tex]\Delta G<0[/tex]

Given, [tex]\Delta H=-310.1\ kJ/mol=-310100\ J/mol[/tex]

[tex]\Delta S=-89.00\ J/K.mol[/tex]

So,

[tex]\Delta H-T\times \Delta S<0[/tex]

Thus, applying values as:-

[tex]-310100-T\times (-89.00)<0[/tex]

So, T = 3484.27 K

The conversion of T( °C) to T(K) is shown below:

T( °C) = T(K) - 273.15

So,

T = (3484.27 - 273.15) K = 3211.12 K

The temperature below which the reaction is spontaneous is:- 3211.12 K