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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

Respuesta :

Answer:

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

Explanation:

For a reaction occurring at constant pressure (P)-

                             [tex]\Delta H_{rxn}=\Delta U_{rxn}+P\Delta V[/tex]

where [tex]\Delta H[/tex], [tex]\Delta U[/tex] and [tex]\Delta V[/tex] represent change in enthalpy, change in internal energy and change in volume respectively. [tex]P\Delta V[/tex] 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 [tex]\Delta H_{rxn}[/tex] = -162. kJ, [tex]P\Delta V[/tex] = -194. kJ

So [tex]\Delta U_{rxn}=\Delta H_{rxn}-P\Delta V[/tex]

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

                = 32.0 kJ