#1. What is an isolated system? Do isolated systems exist?
#2. Define the system and the surroundings. If you want to analyze the energy of a water bottle, what would you define as the system and surroundings?
#3. What does the first law of thermodyamics state?
#4. What are two definitions of enthalpy? Give the mathematical definition and the definition under constant pressure.
#5. What is the SI unit of energy? List 3 types of energy.
#6. What energy does the standard heat of formation (`DeltaH_f`) of a compound correspond to? What are the units of the heats of formation?
#7. What is the standard state of an element?
#8. What is the specific heat of a compound? What is unique about the specific heat of water?
#9. What is a state function?
#1. A gas at `3 "atm"` from `1.5 "L"` to `2 "L"`. At the same time, the gas releases 300J of energy in the form of heat. What is the total change in energy for the gas? Is this process exothermic or endothermic?
#2. 0.75g of `CH_4O_((l))` is placed into a calorimeter consisting of 2000g of water at 30°C. What is the final temperature of the water after all of the `CH_4O_((l))` has been combusted? The heat of combustion for `CH_4O_((l))` is `(55500 kJ)/"mol"`.
#3. Calculate the change in enthalpy for the reaction
`PCl_(5(g)) rArr PCl_(3(g)) + Cl_(2(g))`
by using the following reactions:
`(1) 4PCl_(3(g)) rArr P_(4(s)) + 6Cl_(2(g)) : DeltaH=2449 kJ`
`(2) 4PCl_(5(g)) rArr P_(4(s)) + 10Cl_(2(g)) : DeltaH=3440 kJ`
#1. An isolated system is one that does not exchange energy or matter with its surroundings. True isolated systems do not exist in the real world, as all systems will exchange some amount of energy or matter. There are systems that are "isolated enough" such that we can treat them as isolated systems, however.
#2. The system is the part of the universe that we are examining. The surroundings are everything outside of the system. In this case, the water bottle would be the system and the rest of the universe would be the surroundings.
#3. The first law states that, for any isolated system, the change in energy of that system is 0. This is often stated as the "conservation of energy."
#4. The mathematical definition of enthalpy is `DeltaH=DeltaU +Delta(PV)`. In situations where the pressure is constant, the equation is simplified to `DeltaH=q`.
#5. The SI unit of energy is the Joule (`J`). Sometimes it is given in `L*"atm"`, where the conversion factor is `(101.3 J)/(L*"atm")`. The three types of energy discussed in this section are heat `(q)`, work `(w)`, and enthalpy `(H)`.
#6. The standard heat of formation corresponds to the energy required to form 1 mole of that compound from its constituent elements in their standard states. This also corresponds to the energy necessary to break and reform bonds. Heats of formation are given in units of `J/("mol")`.
#7. The standard state of an element is the state of matter that the element is in at `25°C` and `1 "atm"`, also known as Standard Temperature and Pressure (STP).
#8. The specific heat of a substance is the amount of heat necessary to heat up one gram of the substance by `1°C`. Water has the highest specific heat of any known liquid, allowing it to be used for thermal regulation.
#9. A state function is a property of a system that only depends on the initial and final states.
#1. This problem is a first law problem. Recall that `DeltaU =q+w`. All that we have to do is calculate `q` and `w`.
`q` is easy as we're given it. Since the gas is releasing heat, the process is exothermic. This means that the sign is negative. `q= -300 J`
`w=-pDeltaV=-(3 "atm")(2L-1.5L)=-(3 "atm")(0.5L)=-1.5 L*"atm"=152 J`
Now that we know `q` and `w`, we can just add them together according to the first law.
`DeltaU=(-300 J)+(152 J)=-148 J`
Since energy is being released from the system into the surroundings, the process is exothermic.
#2. This problem requires application of both the first law and calorimetry. We can consider the calorimeter an ideal system and thus, `DeltaU=0.` This means that `q_"combustion"+q_"water"=0`.
Since we want to find the `DeltaT` of the water, we can expand this expression.
`q_"combustion"=-q_"water"`
`n_(CH_4O)*DeltaH_"combustion"=-m_"water"c_"water"DeltaT`
where `n_(CH_4O)` is the number of moles `CH_4O`, and `DeltaH_"combustion"` is the enthalpy from combustion.
`-(n_(CH_4O)*DeltaH_"combustion")/(m_"water"c_"water")=DeltaT`
`-(n_(CH_4O)*DeltaH_"combustion")/(m_"water"c_"water")=T_f-T_i`
`-(n_(CH_4O)*DeltaH_"combustion")/(m_"water"c_"water")+T_i=T_f`
The molecular weight of `CH_4O` is `(32 g)/"mol"`. After converting to moles, we have `n_(CH_4O)=0.0234 "mol"` Plugging in values, we get
`T_f= -((0.0234 "mol")((-55500*10^3 J)/"mol"))/((2000 g)((4.184 J)/g*K))+(303 K)`
Notice that the heat of combustion is negative. This is because combusting a molecule releases heat.
`T_f=457 K=184°C`
#3. Use Hess's Law. Some key things to notice:
`PCl_5` is on the reactants side.
`P_4` does not exist in the final reaction.
`PCl_3` and `Cl_2` will both end up on the reactants side
There is only 1 mole of `PCl_5` in the final equation whereas there are 4 in the constituent equations.
First, let's flip (1) so that we get `PCl_3` on the products side. Notice that this also accomplishes cancelling out the `P_4`.
`(1) P_(4(s)) + 6Cl_(2(g)) rArr 4PCl_(3(g)) : DeltaH=-(2449 kJ)`
`(2) 4PCl_(5(g)) rArr P_(4(s)) + 10Cl_(2(g)) : DeltaH=3440 kJ`
Since we want a final equation with only 1 mole of `PCl_5`, we can divide (2) by 4. By doing so, the `P_4` are now unbalanced, so we may as well divide (1) by 4 as well.
`(1) 1/4 P_(4(s)) + 6/4 Cl_(2(g)) rArr PCl_(3(g)) : DeltaH=-1/4 (2449 kJ)`
`(2) PCl_(5(g)) rArr 1/4 P_(4(s)) + 10/4 Cl_(2(g)) : DeltaH=1/4 (3440 kJ)`
Now if we add everything up, we should get the original equation.
`1/4 P_(4(s)) + 6/4 Cl_(2(g))+PCl_(5(g))rArr PCl_(3(g)) +1/4 P_(4(s)) + 10/4 Cl_(2(g))`
After cancelling out the reactants on both sides, we get the original equation.
`PCl_(5(g)) rArr PCl_(3(g)) + Cl_(2(g))`
`DeltaH_"rxn"=-1/4 (2449 kJ)+1/4 (3440 kJ)`
`DeltaH_"rxn"=247 kJ`