Answer:
2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g ) and A ( s ) + B ( s ) ⟶ C ( g )
Explanation:
The definition of work, W , for a constant pressure process in thermodynamics is:
W = -PΔV
where P is the constant pressure, and ΔV is the change in volume.
We know also from the ideal gas law that
PV = nRT
Therefore
PΔV = - Δn (rxn) RT
where
Δn (rxn) = number of moles gas produced in the reaction - number of moles of gas reactants.
So what we have to do to answer this question is determine the Δn (rxn) and plug it in our equation to check for the sign.
if resultant work is positive, work is done by the surrounding on the system.
If resultant work is negative, work is done by the system on the surroundings.
For 2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )
Δn (rxn) = 3 - 4 = -1 ( W = - (- 1)RT = + RT )
Work is positive, the work is done by the surroundings on the system.
For 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )
Δn (rxn) = 5 - 2 - 2 = + 1 ⇒ W = - ( + 1 )RT = - RT
The work is negative, hence work is done by the system on the sorroundings.
For A ( g ) + 2 B ( g ) ⟶ 2 C ( g )
Δn (rxn) = 2 - 1 - 2 = - 1 ( W = - (-1)RT = +RT )
Work is positive, hence work is not done on the surroundings
For A ( s ) + B ( s ) ⟶ C ( g )
Δn (rxn) = 1 ( W = - (+1)RT = -RT )
Work is negative, hence work is done on the surroundings by the system
