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Question: Suppose that a reaction has \(\text{ }\Delta\text{ H = 40 kJ}\)and\(\text{ }\Delta\text{ S = 50 kJ/K...

Suppose that a reaction has  Δ H = 40 kJ\text{ }\Delta\text{ H = 40 kJ}and Δ S = 50 kJ/K\text{ }\Delta\text{ S = 50 kJ/K}. At what temperature range it will change from non-spontaneous to spontaneous?
A.0.8 K\text{0}\text{.8 K}to 1.0 K\text{1}\text{.0 K}
B.799 K\text{799 K}to800 K\text{800 K}
C.800 K\text{800 K}to 801 K\text{801 K}
D.799 K\text{799 K}to 801 K\text{801 K}

Explanation

Solution

The spontaneity of a reaction is decided by the Gibbs free energy of a reaction. If the Gibbs free energy has a negative value then the reaction proceeds with spontaneity, while if the reaction has positive value for Gibbs free energy then the reaction is nonspontaneous.

Complete step by step answer:
The mathematical presentation of the Gibbs free energy G = H - TS\text{G = H - TS}
The free energy change of a reaction can be defined as, the change in the enthalpy of the reaction ( Δ H)\left( \text{ }\Delta\text{ H} \right)minus the product of the temperature (Kelvin) and change in the entropy of the system ( Δ S)\left( \text{ }\Delta\text{ S} \right)
ΔG = ΔH - TΔS\Delta \text{G = }\Delta \text{H - T}\Delta \text{S}
Here,  Δ H = 40 kJ\text{ }\Delta\text{ H = 40 kJ} and  Δ S = 50 kJ/K\text{ }\Delta\text{ S = 50 kJ/K}
Now as said earlier,
If ΔG \Delta \text{G }< 0, the process is a spontaneous one,
If ΔG \Delta \text{G }> 0, the process is a non-spontaneous one,
ΔG \Delta \text{G }= 0, the process is at equilibrium.
ΔG = 40 - T×50\Delta \text{G = }40\text{ - T}\times 50
If the temperature is0.8 K\text{0}\text{.8 K}, ΔG = 40 - (0.8×50)=0\Delta \text{G = }40\text{ - }\left( 0.8\times 50 \right)=0the system is at equilibrium, while at 1.0 K\text{1}\text{.0 K}, ΔG = 40 - (1×50)=−10\Delta \text{G = }40\text{ - }\left( 1\times 50 \right)=-10kJ, the process is spontaneous.
If the temperature is799 K\text{799 K}, ΔG = 40 - (799×50)=−39,910\Delta \text{G = }40\text{ - }\left( 799\times 50 \right)=-39,910the system is at equilibrium, while at 800.0 K\text{800}\text{.0 K}, ΔG = 40 - (800×50)=−39,960\Delta \text{G = }40\text{ - }\left( 800\times 50 \right)=-39,960kJ, the process is spontaneous.
If the temperature is 799 K\text{799 K}, ΔG = 40 - (799×50)=−39,910\Delta \text{G = }40\text{ - }\left( 799\times 50 \right)=-39,910the process is spontaneous, while at 800.0 K\text{800}\text{.0 K}, ΔG = 40 - (800×50)=−39,960\Delta \text{G = }40\text{ - }\left( 800\times 50 \right)=-39,960kJ, the process is spontaneous.
If the temperature is 801 K\text{801 K}, ΔG = 40 - (801×50)=−40,010\Delta \text{G = }40\text{ - }\left( 801\times 50 \right)=-40,010, the process is spontaneous.
Among all the processes,

The one in which the process will become spontaneous from a non-spontaneous one is option A, 0.8 K\text{0}\text{.8 K}to 1.0 K\text{1}\text{.0 K}, which is the correct Option.
Note:
The Gibbs free energy is defined as the energy change that is associated with a chemical change that is used to do work. The free energy of a system is the sum of the enthalpy of the system (H) and the product of the temperature (Kelvin) and the entropy of the system (S).