Question

For a particular reaction, AH = -16.1 kJ/mol and AS = -21.8 J/(mol-K). Assuming these values change very little with temperature, at what temperature does the reaction change from no to spontaneous in the forward direction? 5 T= Is the reaction in the forward direction spontaneous at temperatures s greater than or less than the calculated temperature? less than O greater than

Answer 1

Answer:

Step 1: Explanation

Gibbs free energy of a system is defined as the enthalpy of the system minus the product of the temperature times the entropy of the system.The energy is associated with a chemical reaction that can be used to do work.

ΔG = ΔH−TΔS

where,

ΔG = Gibbs Free energy

ΔH = Enthalpy of the system

ΔS = Entropy of the system

T = temperature

Note:

Case 1: A reaction is spontaneous if the change in gibbs free energy (ΔG) is negative

Case 2: A reaction is non spontaneous if the change in gibbs free energy (ΔG) is positive

Step 2: Solution

Given,

ΔH = -16.1 kJ/mol = -16100 J/mol [ note: 1 kJ =1000 J ]

ΔS = -21.8 J/(mol.K)

T = we need to calculate

We know for the reaction to be spontaneous gibbs free energy (ΔG) must be negative ( less than 0)

ΔG = ΔH−TΔS

=> ΔG = ΔH−TΔS < 0

=> ΔH−TΔS < 0

=> T >  ΔH / ΔS

=> T > -16100 J/mol  / -21.8 J/(mol.K)

=>  T > 738.532 K

Answer:

(a) The temperature at which the reaction becomes spontaneous = 738.532 K

(b) The reaction will be spontaneous when temperature = greater than