Question

For a certain chemical reaction, ΔH°=−34.2 kJ and ΔS° = −88.5 J/K.

Does the reaction lead to an increase or decrease in the disorder of the system?
O increase
O decrease

Answer 1

The question asks for the calculation of the standard free-energy change (ΔG°) using the provided standard enthalpy change (ΔH°) and standard entropy change (ΔS°), indicating whether the reaction is exothermic and spontaneous under standard conditions.

Step-by-step explanation:

The student's question pertains to the calculation of the standard free-energy change (ΔG°) for a chemical reaction, given the standard enthalpy change (ΔH°) and the standard entropy change (ΔS°). To calculate ΔG°, the formula ΔG° = ΔH° - TΔS° is used, where T represents the temperature in Kelvin.

It's important to remember that ΔH° and ΔS° should be in compatible units, with enthalpy typically given in kJ/mol and entropy in J/(mol·K). The given values ΔH° is -34.2 kJ and ΔS° is -88.5 J/K; a negative ΔH° indicates that the reaction is exothermic, while a negative ΔS° indicates a decrease in disorder. By incorporating these values into the equation, one can determine if the reaction tends to be spontaneous (ΔG° is negative) under standard conditions at a given temperature.