in the equation 2SO2 + O2 -> 2SO3

how do you find if work done is >, <, or = to 0?

The system has 3 mols gas on the left and 2 on the right; therefore, the system decreases in volume which means work is done ON the system if the pressure is constant. Work done ON the system is + (or > 0). Check my thinking.

To determine if the work done in the given reaction is greater than (>), less than (<), or equal to (=) zero, you need to calculate the change in Gibbs free energy (∆G) for the reaction.

The Gibbs free energy change can be calculated using the equation:

∆G = ∆H - T∆S

Where:
- ∆H is the change in enthalpy (heat) of the system
- T is the temperature in Kelvin
- ∆S is the change in entropy of the system

If ∆G is negative (∆G < 0), the work done is greater than zero, indicating that the reaction is spontaneous in the forward direction. This means that the reaction will proceed naturally without the need for any external energy input.

If ∆G is positive (∆G > 0), the work done is less than zero, indicating that the reaction is not spontaneous in the forward direction. The reaction would require an input of energy to proceed.

If ∆G is zero (∆G = 0), the work done is equal to zero, indicating that the reaction is at equilibrium. In this case, the forward and reverse reactions occur at the same rate, and there is no overall change.

To calculate ∆G, you need to know the enthalpy change (∆H) and the entropy change (∆S). The values for ∆H and ∆S can be obtained from experimental data or calculated using standard enthalpies of formation (∆H°f) and standard entropies (∆S°) for the reactants and products.

Once you have the values for ∆H and ∆S, you can plug them into the equation ∆G = ∆H - T∆S and calculate the value of ∆G to determine if the work done is greater than (>), less than (<), or equal to (=) zero.