Question

measurements show that the energy of a mixture of gaseous reactants decreases by 201. 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 167. kj of work is done on the mixture during the reaction.
calculate the change in enthalpy of the gas mixture during the reaction. round your answer to 3 significant digits.
is the reaction exothermic or endothermic?
options: exothermic, endothermic

Explanation:

Step1: Recall enthalpy change definition

At constant pressure, $\Delta H = q_p$. Also, $\Delta U = q + w$. Here, $\Delta U$ is the energy change of the system (reactants: decreases by 201 kJ → $\Delta U = -201$ kJ). Work done on the system: $w = +167$ kJ.

Step2: Calculate $q_p$ (equal to $\Delta H$)

Rearrange $\Delta U = q + w$ to $q = \Delta U - w$. Substitute values: $q = -201$ kJ $- 167$ kJ $= -368$ kJ. Since pressure is constant, $\Delta H = q_p = -368$ kJ.

Step3: Determine reaction type

Negative $\Delta H$ means heat is released → exothermic.

Answer:

• Change in enthalpy: -368 kJ
• Reaction type: exothermic