Calculate the work involved if a reaction with an enthalpy change of -2418 kJ is carried out in a vessel with a mobile, frictionless piston. Other details: the reaction is H2(g) + 1/2Oxygen2(g) yields H2O(g) with enthalpy change of -241.8 kJ/mol. The product is 180.16gH2O=10.000 mol H2O (which is how I got the enthalpy of reaction).
I know delta E is close to delta H, so using delta H = delta E + PdeltaV, I get PdeltaV = 0. I also know that work = -PdeltaV, but that would mean that 0 work was done. I know that this is logically impossible, because the piston would have to move. What equation(s) should I use?
To calculate the work involved in the reaction, you can use the equation:
Work = -P × ΔV
In this case, since the vessel has a mobile, frictionless piston, the pressure-volume work can be expressed as:
Work = -ΔnRT
where Δn is the change in the number of moles of gas, R is the ideal gas constant, and T is the temperature.
To find the change in the number of moles (Δn), you can use stoichiometry based on the balanced equation for the reaction:
H2(g) + 1/2O2(g) → H2O(g)
From the balanced equation, you can see that 1 mole of H2 produces 1 mole of H2O. Therefore, for a reaction producing 10.000 moles of H2O, the change in the number of moles (Δn) is also 10.000 moles.
Now, you need to determine the value for R and the temperature (T). The value of R depends on the units you are using. In this case, since you're working with kJ and kJ/mol, you should use the value R = 0.0083145 kJ/(mol·K). The temperature (T) needs to be provided in Kelvin (K).
Once you have the values for Δn, R, and T, you can substitute them into the equation for work to calculate the actual work involved in the reaction.
To calculate the work involved in this reaction, you need to consider the equation:
ΔH = ΔE + PΔV
where ΔH is the enthalpy change, ΔE is the change in internal energy, P is the pressure, and ΔV is the change in volume.
Now, in this case, you mentioned that the reaction is taking place in a vessel with a mobile, frictionless piston. This means that the reaction involves a change in volume, and therefore, there is work being done by the system on the surroundings or vice versa.
To calculate the work, you need to determine the change in volume (ΔV) and the pressure (P).
Since the reaction involves gases, you can use the ideal gas law, PV = nRT, to relate pressure, volume, and the number of moles.
Given that the reaction produces 10.000 mol of H2O, and assuming all other gases are ideal, you can calculate the initial and final volumes of the gases using the ideal gas law.
The initial volume can be calculated using the initial number of moles of H2 and O2, and the final volume can be calculated using the final number of moles of H2O.
Once you have the initial and final volumes, you can determine the change in volume (ΔV) by subtracting the initial volume from the final volume.
Now, since the reaction is carried out in a mobile, frictionless piston, the pressure (P) will be constant throughout the process. Therefore, you can use the constant pressure (ΔP = 0) to simplify the equation:
ΔH = ΔE + PΔV
ΔH = ΔE
This means that the enthalpy change (ΔH) is equal to the change in internal energy (ΔE), which indicates that no work is done because there is no change in volume.
So, in this case, the equation you described, work = -PΔV, is not applicable because there is no change in volume (ΔV = 0), and therefore, the work involved is indeed zero.
In summary, for a reaction carried out in a vessel with a mobile, frictionless piston, if there is no change in volume, the work involved is zero.