For the gaseous equilibrium represented below, it is observed that greater amounts of PCl3 and Cl2 are produced as the temperature is increased.
PCl5 (g) <==> PCl3 (g) + Cl2 (g)
a) what is the sign of delta S for the reaction? Provide a legitimate rationale for the correct answer.
b) what change, if any, will occur in delta G° for the reaction as the temperature is increased? Provide a legitimate rationale for the correct answer.
To determine the sign of delta S for the reaction, we can consider the changes in entropy associated with the formation of the products and the reactants.
a) The reaction involves the conversion of PCl5 into PCl3 and Cl2, which means that the number of moles of gas increases. The increase in the number of moles of gas generally leads to an increase in entropy. Therefore, we can expect that the delta S for the reaction is positive.
To calculate the change in Gibbs free energy (delta G°) for the reaction as the temperature increases, we need to consider the relationship between delta G° and temperature.
b) The equation that relates delta G°, delta H° (change in enthalpy), and delta S° (change in entropy) is:
delta G° = delta H° - T * delta S°
where T is the temperature in Kelvin.
Considering the equation, we can see that as the temperature increases, the value of T * delta S° term becomes larger. If delta S° is positive (as we determined in part a), the T * delta S° term will have a positive contribution to delta G°. This means that the change in delta G° will become less negative or possibly even positive.
In other words, as the temperature increases, the change in Gibbs free energy (delta G°) for the reaction becomes less favorable or more positive. Therefore, the reaction becomes less spontaneous as the temperature increases.
It's important to note that while this explanation is based on general trends, actual values and the precise behavior of the reaction would require more specific information such as the actual values of delta H° and delta S° for the reaction.