2HBr(g) + Cl2(g)2HCl(g) + Br2(g)
ΔH° = -81.1 kJ and ΔS° = -1.2 J/K
The maximum amount of work that could be done when 2.05 moles of HBr(g) react at 274 K, 1 atm is...?
The maximum amount of useful work that can be done by a chemical reaction is given by dGo = dHo - TdSo. The units of dGo are kJ/mol so multiply the answr you get by 2.05moles.
Don't forget that dH usually is given in kJ/mol and dS usually is given in J/mol. The usual method is to convert the dSo value to kJ first.
Post your work if you get stuck.
To determine the maximum amount of work that could be done during a chemical reaction, we can use the equation:
ΔG = ΔH - TΔS
where:
ΔG = Gibbs free energy change
ΔH = enthalpy change
T = temperature in Kelvin
ΔS = entropy change
Given:
ΔH° = -81.1 kJ
ΔS° = -1.2 J/K
T = 274 K
First, we need to convert the given values into the appropriate units (kJ and kJ/K):
ΔH = -81.1 kJ = -81,100 J
ΔS = -1.2 J/K
Next, substitute the values into the equation:
ΔG = ΔH - TΔS
ΔG = -81,100 J - (274 K)(-1.2 J/K)
Now, let's calculate the value of ΔG:
ΔG = -81,100 J + 328.8 J
ΔG = -80,771.2 J
Since the pressure is given as 1 atm, we can convert the value of ΔG from joules to liters-atmospheres:
ΔG = -80,771.2 J * (1 L-atm / 101.325 J)
ΔG = -797.0 L-atm
Therefore, the maximum amount of work that could be done during this reaction is -797.0 liters-atmospheres. Note that this value is negative because the reaction is exothermic, meaning it releases energy instead of absorbing it.