The equation given is 2SO2(g)+O2(g)->2SO3(g),

and if 224L/h of O2 (at standard temp and pressure) is consumed by the reaction,

how much power is produced?

I can't go further than P=Work/Time and Work=P times change in V

I've tried 224x2-224x3/60^2 but that's not the answer :( The answer is 550W. No idea how.
Some help would be much appreciated!

To determine the power produced in this reaction, we need to use the ideal gas law and stoichiometry.

The ideal gas law states that 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 are given the volume of O2, which is 224 L. The reaction equation shows that 1 mole of O2 reacts to produce 2 moles of SO3. Therefore, we can find the number of moles of O2 consumed as follows:

n(O2) = (224 L) / (22.4 L/mol) = 10 moles

Since 2 moles of O2 produce 2 moles of SO3, the number of moles of SO3 produced is also 10 moles.

Now, to determine the power produced, we need to calculate the work done in the reaction. The work done is given by the equation W = -PΔV, where P is the pressure and ΔV is the change in volume.

Since the reaction equation shows that 2 moles of SO2 react to produce 2 moles of SO3, the change in volume is zero. Therefore, there is no work done in this process.

As a result, the power produced is zero, not 550 W. It seems there may be a misunderstanding or mistake in the provided answer.

Please recheck the given equation, the expected answer, or any additional information that may have been provided to find the correct solution.