An amount of 414.27 g of zinc sulfide (ZnS, MM=97.44 g/mol) reacts with 99.58 g of oxygen gas (O2, MM=31.999 g/mol) at a constant temperature of 298.0 K and constant pressure of 1.00 atm. The reaction:
2 ZnS(s) + 3 O2(g) --> 2 ZnO(s) + 2 SO2(g)
dH = -818.58 kJ
a) How much heat is released by the reaction?
b) How much work is done by the atmosphere on the reacting mixture for the conditions stated above?
Please help. Thank you so much!
To calculate the heat released by the reaction, we need to use the equation:
q = n * dH
where q is the heat released, n is the number of moles of the limiting reactant, and dH is the enthalpy change for the reaction.
First, let's calculate the number of moles of zinc sulfide (ZnS):
moles of ZnS = mass of ZnS / molar mass of ZnS
moles of ZnS = 414.27 g / 97.44 g/mol
moles of ZnS ≈ 4.25 mol
Since the balanced equation shows that the stoichiometric ratio between ZnS and dH is 2: -818.58 kJ, we need to divide the number of moles of ZnS by 2:
n = 4.25 mol / 2 = 2.125 mol
Now, we can calculate the heat released:
q = n * dH
q = 2.125 mol * -818.58 kJ/mol
q ≈ -1743.85 kJ
Therefore, the heat released by the reaction is approximately -1743.85 kJ.
To calculate the work done by the atmosphere on the reacting mixture, we need to use the equation:
w = -P * ΔV
where w is the work done, P is the pressure, and ΔV is the change in volume.
Since the pressure is given as 1.00 atm and the volume change is not specified, we cannot directly calculate the work done using the given information. The work done depends on the change in volume, which we need to have to proceed with the calculations.
If the question provides the change in volume or any other information related to the change in volume, please let me know so I can assist you further.