Classify the possible combinations of signs for a reaction's ∆H and ∆S values by the resulting spontaneity.

A) Spontaneous as written at all temperatures.

B) Spontaneous in reverse at all temperatures.

C) Spontaneous as written above a certain temperature.

D) Spontaneous as written, below a certain temperature.

1) ∆H = + / / ∆S = +

2) ∆H = - / / ∆S = +

3) ∆H = - / / ∆S = -

4) ∆H = + / / ∆S = -

Correct combinations are

For option A it is 2

For option B it is 4

For option D it is 3

For option C it is 1

Explanation:

For checking the spontaneity of a reaction, we have to check the sign of ΔG using the below formula

ΔG = ΔH - T*ΔS

where

ΔG is the change in Gibbs free energy

ΔH is the change in enthalpy

T is the temperature

ΔS is the change in entropy

For spontaneous reactions, ΔG must be less than zero and for non-spontaneous reactions ΔG must be greater than zero but for an equilibrium reaction ΔG must be equal to zero

So in case of 1 as ΔH and ΔS are positive if the temperature is above a certain value then ΔG will be less than zero

So in case of 2 as ΔH is negative and ΔS is positive then ΔG will always be less than zero at all temperatures

So in case of 3 as ΔH and ΔS are negative if the temperature is below a certain value then ΔG will be less than zero

So in case of 4 as ΔH is positive and ΔS is negative then ΔG will always be greater than zero but in reverse direction as ΔG is less than zero therefore in reverse direction the reaction will be spontaneous at all temperatures