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Given the values of ΔH^{∘}rxn, ΔS^{∘}rxn, and T below, determine Δ Suniv.

A. ΔH^{∘}rxn= -118 kJ , ΔS^{∘}rxn= 258 J/K , T= 294 K

B. ΔH^{∘}rxn = 118 kJ , ΔS^{∘}rxn = -258 J/K , T= 294 K

C. ΔH^{∘}rxn =- 118 kJ , ΔS^{∘}rxn = -258 J/K , T= 294 K .

D. ΔH^{∘}rxn =- 118 kJ , ΔS^{∘}rxn = -258 J/K , T= 545 K .

Predict whether or not the reaction in part A will be spontaneous.

Predict whether or not the reaction in part B will be spontaneous.

Predict whether or not the reaction in part C will be spontaneous.

Predict whether or not the reaction in part D will be spontaneous.

**Entropy of the Universe**

Second law of thermodynamics says that the universe’s entropy will always be greater than zero in a spontaneous system . This implies that the total of the entropy in the system and the environment should be greater than zero in a spontaneous process.

To calculate the value the entropy of the universe using the following general work equation should be used:

ΔS

_{universe }= ΔS_{system }+ ΔS_{surrounding}ΔS

_{universe }= ΔS_{rxn }− ΔH_{rx}_{n}/ T(a) ΔS

_{universe }= 258J/K− (−118kJ×1000J/kJ)/294k = 659.3605J/K(b) ΔS

_{universe }= −258J/K− (118kJ×1000J/kJ)/294k = −659.3606J/K(c) ΔS

_{universe }= −258J/K− (−118kJ×1000J/kJ)/294k = 143.3605J/K(d) ΔS

_{universe}= −258J/K− (−118kJ×1000J/kJ)/545k = −41.4862J/Ka. The change in entropy in the universe is positive, which indicates that the process can be described as

spontaneous.b. The change in entropy in the universe is negative. This means the process is

not-spontaneous.C. A change of the entropy of the universe is positive, which implies that the this process can be described as

spontaneousd. The increase in entropy of the universe is negative. This indicates that the it is

non-spontaneous