The reaction between hydrogen and oxygen to yield water vapor has δh∘=−484kj: 2h2(g)+o2(g)→2h2o(g)δh∘=−484kj

how much pv work is done in kilojoules for the reaction of 3.20 mol of h2 with 1.60 mol of o2 at atmospheric pressure if the volume change is −22.6l?

express your answer using three significant figures

The PV work in kilojoules for the reaction is .

Further Explanation:

First law of thermodynamics:

The law states that energy cannot be created or destroyed but only changes its form and versa.

The internal energy of the system defines the total energy of the system and is sum of enthalpy of the system and work done. Internal energy only depends on the temperature of the system. The mathematical expression of the internal energy is as follows:  

Where,

U is the internal energy of the system.

H is the enthalpy of the system.

W is the work-done.

The formula for workdone by the system is as follows:  

Where,

P is the pressure in atmosphere.

is the change in the volume in liter.

The atmospheric pressure is .

The change in volume is .

Substitute these values in above equation to calculate the work done by the system.

The correct answer in 3 significant figures is .

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Answer Details:

Grade: Senior School

Chapter: Thermodynamics

Subject: Chemistry

Keywords: Internal energy,thermodynamics,PV work, kilojoules, volume change, enthalpy, work done. significant figures,2.29KJ, atmospheric pressure.

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