Help......me.....

Question

For each of the reactions at constant pressure, determine whether the system does work on the surroundings, the surroundings does work on the system, or essentially no work is performed.

Work done by the system

Work done on the system

No work done

CH4(g)+2O2(g)⟶CO2(g)+2H2O(g)

NH3(g)+HCl(g)⟶NH4Cl(s)

NaHCO3(s)+HC2H3O2(aq)⟶NaC2H3O2(aq)+H2O(l)+CO2(g)

2H2O2(g)⟶O2(g)+2H2O(g)

Answer 1

When the system is at constant pressure, we have the following relationship:

work = -pressure x change in volume

w = -P∆V (note the minus sign in front of the P∆V)

By convention,

if w is negative, the system is doing work on the surroundings i.e. work done by the system

if w is positive, the surrounding is doing work on the system, i.e. work done on the system

if w is zero, no work is done.

CH4(g)+2O2(g)⟶CO2(g)+2H2O(g)

3 moles gas on left and 3 moles gas on right. No change in volume. No work done.

NH3(g)+HCl(g)⟶NH4Cl(s)

2 moles gas on left and no moles gas on right. Volume will decrease making -P∆V positive. Work done on the system

NaHCO3(s)+HC2H3O2(aq)⟶NaC2H3O2(aq)+H2O(l)+CO2(g)

no moles of gas on the left and 1 mole gas on the right. Volume will increase making -P∆V negative. Work done by the system

2H2O2(g)⟶O2(g)+2H2O(g)

2 moles gas on the left and 3 moles gas on the right. Volume will increase making -P∆V negative. Work done by the system.