Thermochemistry Practice Questions & Answers

OpenStax 5.1 defines energy as 'the capacity to supply heat or do work.' Work is described as causing matter to move against an opposing force.

Using the first law of thermodynamics: $\Delta U = q + w$ΔU=q+w. Here, heat is absorbed ($q = +300\ \text{J}$q=+300 J) and work is done on the system ($w = +100\ \text{J}$w=+100 J). Thus, $\Delta U = 300 + 100 = +400\ \text{J}$ΔU=300+100=+400 J.

According to the sign convention in OpenStax 5.3, work ($w​$w​) is negative if it is done by the system. A gas expanding pushes against the surroundings, doing work on them, so $w < 0$w<0.

OpenStax 5.1 states: 'Thermal energy is kinetic energy associated with the random motion of atoms and molecules.' Temperature is a quantitative measure of hot or cold.

OpenStax 5.3 explains that for processes at constant pressure (typical for open containers), $\Delta H = q_p$ΔH=qp​. This makes it convenient because the heat measured directly equals $\Delta H$ΔH.

OpenStax 5.3 explicitly states: 'Enthalpy is also a state function... whereas heat and work are not state functions.'

From LibreTexts 5.3 and OpenStax 5.3, $\Delta H = \Delta U + \Delta(PV)$ΔH=ΔU+Δ(PV). For ideal gases at constant temperature, $\Delta(PV) = \Delta(nRT) = RT\Delta n$Δ(PV)=Δ(nRT)=RTΔn. Therefore, $\Delta H - \Delta U = \Delta nRT$ΔH−ΔU=ΔnRT.

LibreTexts 5.3 states: 'If the reaction occurs in a closed vessel, the volume of the system is fixed... the heat flow ($q_v$qv​) must equal $\Delta U$ΔU.'

OpenStax 5.3 states: 'A negative value of an enthalpy change, $\Delta H < 0$ΔH<0, indicates an exothermic reaction.'

In an endothermic reaction, heat is absorbed. This means the system gains energy, placing the products at a higher potential energy level than the reactants (OpenStax 5.1/5.3).