Question

Measurements show that the enthalpy of a mixture of gaseous reactants decreases by 162. kJ during a certain chemical reaction, which is carried out at a constant pressure. Furthermore, by carefully monitoring the volume change it is determined that -194. kJ of work is done on the mixture during the reaction. Calculate the change in energy of the gas mixture during the reaction. Round your answer to 3 significant digits

Answer 1

Change in energy of the gas mixture during the reaction is 32.0 kJ.

Step-by-step explanation:

For a reaction occurring at constant pressure (P)-

`\Delta H_(rxn)=\Delta U_(rxn)+P\Delta V`

where
`\Delta H`,
`\Delta U` and
`\Delta V` represent change in enthalpy, change in internal energy and change in volume respectively.
`P\Delta V` represents work done at constant pressure.

Here change in energy of the gas mixture actually indicates change in internal energy of the gaseous mixture.

Here
`\Delta H_(rxn)` = -162. kJ,
`P\Delta V` = -194. kJ

So
`\Delta U_(rxn)=\Delta H_(rxn)-P\Delta V`

= (-162. kJ) - (-194. kJ)

= 32.0 kJ