should the 2.2 kJ be a negative or positive?

when are we suppose to write the minus in the answer?

Calculate the work done when 50 g of iron reacts with hydrochloric acid to produce FeCl2(aq) and hydrogen in (a) a closed vessel of fixed volume. (b) an open beaker at 25C. Collect your thoughts You need to judge the magnitude of the volume change and then to decide how the process occurs. If there is no change in volume, there is no expansion work however the process takes place. If the system expands against a constant external pressure, the work can be calculated from eqn 21.6. A general feature of processes in which a condensed phase changes into a gas is that you can usually neglect the volume of a condensed phase relative to the volume of the gas it forms. The solution In (a) the volume cannot change, so no expansion work is done and w = 0. In (b) the gas drives back the atmosphere and therefore w=-PxAV. The initial volume can be neglected because the final volume (after the production of gas) is so much larger and AV = V-V, V = nRTIpex, where n is the amount of produced. Therefore, ART W---PAV --Pax -ART Pa Because the reaction is Fe(s) + 2 HCl(aq) FeCl(aq) + H2(g), 1 mol H, is generated when 1 mol Fe is consumed, and n can be taken as the amount of Fe atoms that react. Because the molar mass of Fe is 55.85 g moll, it follows that w 150g 55.85g mol- -X(8.3145JK- mot')x(298K) --22k The system (the reaction mixture) does 2.2 kJ of work driving back the atmosphere. Comment. The magnitude of the external pressure does not affect the final result: the lower the pressure, the larger is the volume occupied by the gas, so the effects cancel. Self-test 2A.1 Calculate the expansion work done when 50 g of water is electrolysed under constant pressure at 25C Answer: -10kJ