A reaction will be spontaneous only at low temperatures if both ΔH and ΔS are negative. For a reaction in which ΔH = −320.1 kJ/mol and ΔS = −86.00 J/K · mol.

Determine the temperature (in °C) below which the reaction is spontaneous.

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thomasdecabooter thomasdecabooter · Answer 1 · 2019-12-03T16:44:19+01:00

Answer:

This reaction is spontaneous for temperature lower than 3722.1 Kelvin or 3448.95 °C

Explanation:

Step 1: Data given

ΔH = −320.1 kJ/mol

ΔS = −86.00 J/K · mol.

Step 2: Calculate the temperature

ΔG<0 = spontaneous

ΔG= ΔH - TΔS

ΔH - TΔS <0

-320100 - T*(-86) <0

-320100 +86T < 0

-320100 < -86T

320100/86 > T

3722.1 > T

The temperature should be lower than 3722.1 Kelvin (= 3448.9 °C)

We can prove this with Temperature T = 3730 K

-320100 -3730*(-86) <0

-320100 + 320780 = 680 this is greater than 0 so it's non spontaneous

T = 3700 K

-320100 -3700*(-86) <0

-320100 + 318200 = -1900 this is lower than 0 so it's spontaneous

The temperature is quite high because of the big difference between ΔH and ΔS.

This reaction is spontaneous for temperature lower than 3722.1 Kelvin or 3448.95 °C