Which reaction below is spontaneous at SATP?

∆Hº=34.1 kJ and ∆Sº=46.2 J/K

∆Hº=71.7 kJ and ∆Sº=-4.25 J/K

∆Hº=-11.1 kJ and ∆Sº=-61.8 J/K

∆Hº=9.95 kJ and ∆Sº=57.4 J/K

When ∆G is positive, a chemical reaction is

faster

spontaneous

slower

non-spontaneous

### Step 1 of 2

Given data: ∆Hº=34.1 kJ and ∆Sº=46.2 J/K

∆Hº=71.7 kJ and ∆Sº=-4.25 J/K

∆Hº=-11.1 kJ and ∆Sº=-61.8 J/K

∆Hº=9.95 kJ and ∆Sº=57.4 J/K

Find: When ∆G is positive, a chemical reaction is faster, spontaneous, slower or non-spontaneous.

**Explanation:**

we can use the Gibbs free energy (Δ*G*) relationship:

Δ*G*=Δ*H*−*T*Δ*S*

### Step 2 of 2

To determine spontaneity, we can use the Gibbs free energy (Δ*G*) relationship:

Δ*G*=Δ*H*−*T*Δ*S*

where:

- Δ
*G*is the Gibbs free energy change, - Δ
*H*is the enthalpy change, - Δ
*S*is the entropy change, *T*is the temperature in Kelvin.

For a reaction to be spontaneous at a given temperature, Δ*G* must be negative.

Now let’s analyze each reaction:

- Δ
*H*∘=34.1kJ and Δ*S*∘=46.2J/K Δ*G*=34.1−*T*(0.0462) - Δ
*H*∘=71.7kJ and Δ*S*∘=−4.25J/K Δ*G*=71.7−*T*(−0.00425) - Δ
*H*∘=−11.1kJ and Δ*S*∘=−61.8J/K Δ*G*=−11.1−*T*(−0.0618) - Δ
*H*∘=9.95kJ and Δ*S*∘=57.4J/K Δ*G*=9.95−*T*(0.0574)

**Explanation:**

For spontaneity, we want Δ*G*<0. Check the sign of Δ*G* for each reaction based on the given values of Δ*H* and Δ*S*.

##### Final solution

Additionally, when Δ*G* is positive, a chemical reaction is **non-spontaneous**.

Therefore, the spontaneous reaction at SATP is the one where Δ*G* is negative. Identify the reaction with Δ*G*<0.