For the methanol combustion reaction

2CH3OH(â„“) + 3O2(g) ----> 2CO2(g) + 4H2O(g)
estimate the amount of P [change]V work done and tell whether the work was done on or by the system. Assume a temperature of 27â—¦C.

1. 2.5 kJ, work done by the system
2. 2.5 kJ, work done on the system
3. No work is done in this reaction.
4. 7.5 kJ, work done by the system
5. 7.5 kJ, work done on the system

To estimate the amount of work done in the methanol combustion reaction, we need to consider the change in the number of moles of gas molecules before and after the reaction.

Looking at the balanced equation:
2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(g)

We see that the number of moles of gas molecules decreases by 2 moles on the left side (3 moles of O2) and increases by 6 moles on the right side (2 moles of CO2 and 4 moles of H2O).

In order to calculate the work done, we can use the formula:
Work = -PΔV

Since the volume is not given, we can assume that the reaction takes place in a closed container where the volume remains constant. Therefore, the change in volume (ΔV) is zero.

When ΔV is zero, the work done is also zero. This means that no work is done in this reaction (option 3: No work is done in this reaction).

So the correct answer is option 3: No work is done in this reaction.