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?

I originally posted as Mark; I'm assuming that I was overlooked or no one wants to answer the question. Any help would be appreciated. Thanks!

Apologies for the oversight, Mark. Let's look at the problem and address your concerns step-by-step.

The given reaction is: H2(g) + (1/2)O2(g) → H2O(g) with an enthalpy change of ΔH = -2418 kJ.

To calculate the work involved, we need to consider the change in volume associated with the reaction. Since the reaction involves gases, we can assume that the reaction will cause a change in volume inside the vessel due to the formation of water vapor.

The equation you mentioned, ΔH = ΔE + PΔV, is the correct equation to use, where ΔE represents the change in internal energy of the system, P represents the external pressure, and ΔV represents the change in volume.

You correctly observed that ΔH is approximately equal to ΔE because the reaction is carried out at constant pressure (assuming the external pressure remains constant).

However, in your calculation, you assumed that PΔV = 0. This is not accurate because work should indeed be involved. The piston would need to move to accommodate the change in volume.

To determine the work done, we can use the equation:

Work (W) = -PΔV,

where P is the pressure and ΔV is the change in volume.

To calculate the work involved, we need to determine the change in volume associated with the reaction. To do this, we can use stoichiometry and the given mass of the product.

Given: 180.16 g H2O = 10.000 mol H2O.

From the balanced equation, we can see that 1 mole of H2O(g) is formed. So, for 10.000 mol of H2O, the reaction produces a mole ratio of 10.000 mol H2O : 10.000 mol H2O.

Assuming ideal gas behavior, the change in volume (ΔV) can be calculated using the ideal gas law equation:

PV = nRT,

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

In this case, we're assuming constant temperature (since it is not given), so we can rearrange the equation as:

ΔV = (n2 - n1)RT/P,

where n2 is the final number of moles (10.000 mol), n1 is the initial number of moles (0 mol, assuming no reactants are initially present), R is the ideal gas constant (0.0821 L·atm/mol·K), and P is the pressure.

Note: Make sure to use consistent units for pressure (atm) and volume (L) in the calculation.

Once you calculate the ΔV, the work (W) can be obtained by multiplying ΔV by the pressure (P). Since the problem mentions a frictionless piston and there is no external pressure mentioned, we can assume atmospheric pressure (1 atm) or state it explicitly as given in the problem.

I hope the step-by-step approach helps you with your calculation. Feel free to ask if you have any further questions or clarifications!

I apologize for the oversight. I'm here to help you with your question. In this case, you have correctly identified that the correct equation to use is:

ΔH = ΔE + PΔV

where ΔH is the enthalpy change, ΔE is the internal energy change, P is the pressure, and ΔV is the change in volume.

Since the problem states that the reaction takes place in a vessel with a mobile, frictionless piston, we can assume that the volume can change (ΔV ≠ 0). However, the problem does not provide any information about the change in volume or the pressure.

To calculate the work done in this reaction, we need to find the product of the pressure and the change in volume (PΔV). Since we don't have any information about the pressure or the change in volume, we cannot determine the work done.

It is worth noting that the expression ΔH = ΔE + PΔV represents the First Law of Thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. In this case, if there is no change in volume (ΔV = 0), then the work done (PΔV) will be zero, and the enthalpy change (ΔH) will be equal to the internal energy change (ΔE).

I apologize if the lack of information provided in the problem makes it difficult to calculate the work done. If you have any additional information or specific values for pressure or volume change, please provide them, and I will be happy to assist you further.