Question

Measurements show that enthalpy of a mixture of gaseous reactants decreases by 228. 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 -55kJ of work is done on the mixture during the reaction.
Calculate the change in energy of the gas mixture during the reaction.

Is the reaction exothermic or endothermic?

Answer 1

Total change in energy = Heat produced by system + work done on the system
= 228 - 55
= 173 kJ

Because this energy is released as heat, the reaction is exothermic.

Answer 2

Answer:

-283 KJ, exothermic

Explanation:

We are given that

Enthalpy of a mixture of a gases reactants

Change in enthalpy=$\Delat H$=-228 KJ

Where negative sign represents the enthalpy decreases.

Pressure=Constant.

Work don=w=-55 KJ

We have to calculate the change in energy of the gas mixture during the reaction.

At constant pressure, the change in enthalpy

$\Delta H=\Delta U+P\Delta V$

Where w=$-P\Delta V$

Where P=Constant

$\Delta V=$ Change in volume

$\Delta U=$ Change in energy

$\Delta U=\Delta H-P\Delta V$

$\Delta U=\Delta H+w$

Substitute the values then we get

$\Delta U=-228-55=-283$ KJ

Hence, the change in energy of the gas mixture during the reaction=-283 KJ

Change in enthalpy is negative it means the reaction is exothermic because the energy is evolved.