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 solve this problem, we need to use the concept of stoichiometry and the heat of reaction, as well as the ideal gas law to find the answers. Let's break it down step by step:
a) How much heat is released by the reaction?
First, we need to determine the limiting reactant. To do this, compare the number of moles of each reactant to the stoichiometric coefficients in the balanced equation.
- Zinc Sulfide (ZnS):
414.27 g * (1 mol / 97.44 g) = 4.249 mol
- Oxygen Gas (O2):
99.58 g * (1 mol / 31.999 g) = 3.112 mol
According to the balanced equation, the ratio of ZnS to O2 is 2:3. So, to use up all ZnS, we need 2/3 times the amount of O2.
Required moles of O2 = 2/3 * 4.249 mol = 2.833 mol
Since we have only 3.112 mol of O2, it is in excess, and ZnS is the limiting reactant.
Next, we need to calculate the heat released using the heat of reaction provided:
Q = dH * n
where dH = -818.58 kJ (given) and n is the number of moles of the limiting reactant.
Q = -818.58 kJ * 4.249 mol = -3479.2 kJ
Therefore, the heat released by the reaction is -3479.2 kJ (negative sign indicates exothermic release).
b) How much work is done by the atmosphere on the reacting mixture for the conditions stated above?
To calculate the work done, we need the equation for work done by a gas:
W = -PΔV
where W is the work done, P is the pressure, and ΔV is the change in volume.
Given that the pressure is 1.00 atm, and the reaction is carried out at constant pressure, we assume that ΔV is the same as the volume of gases produced according to the balanced equation.
From the balanced equation, 2 moles of ZnS produce 2 moles of SO2 gas. Therefore, for 4.249 moles of ZnS, we have 4.249 moles of SO2 gas.
Using the ideal gas law, PV = nRT, we can find the volume of SO2 gas produced:
V = (n * R * T) / P
where n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), T is the temperature in Kelvin, and P is the pressure.
V = (4.249 mol * 0.0821 L·atm/mol·K * 298.0 K) / 1.00 atm = 105.07 L
Now, we can calculate the work done:
W = -PΔV
W = -1.00 atm * 105.07 L = -105.07 L·atm
Therefore, the work done by the atmosphere on the reacting mixture is approximately -105.07 L·atm (negative sign indicates work is done on the system).
Remember to double-check the calculations and pay attention to significant figures.