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Thermochemistry
Types of
Problems
Specific Heat q = mcΔT where ΔT = Tf – Ti Use when you are given three of the following: Mass (g), Volume (mL) and a density (g mL-1) , specific heat (J g-1^ °C-1) , change in temperature (K or °C), heat (J)
Calorimetry qrxn + qsol’n + qcal = 0 qrxn = -(qsol’n + qcal) where qsol’n = msol’ncsol’nΔTsol’n and qcal = ccalΔTcal
Use ONLY when the problem states that the reaction occurred in a calorimeter!
Thermochemical Equations A balanced chemical equation (with STATES) + an enthalpy of reaction Some rules for these equations:
- The forward reaction and the reverse reaction have the same magnitude (i.e. number) but opposite signs.
- If you multiply the coefficients in the balanced chemical equation by some number, you MUST also multiply the enthalpy by that number!
Standard Molar Enthalpy of Formation ΔHrxn = Σ[moles Product * ΔHfº(Products)] – Σ[moles Reactant * ΔHfº(Reactants)] Use when you are asked to calculate the enthalpy of the reaction and you have ONLY one balanced equation! Another major hint: the problem will ask you to look at an Appendix
Hess’s Law When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. Use when you are given multiple thermochemical equations and are asked to calculate the enthalpy of a “goal” equation
Enthalpy is a state function! ΔHoverall = ΔHf - ΔHi
Thermochemistry Background Information
Thermodynamics – study of heat and its transformations
- Thermochemistry – branch of thermodynamics that deals with heat involved in chemical reactions
Defining the change in heat: System – the part of the universe whose change we are going to observe Surroundings – everything else relevant to the change we are going to observe
ΔE = E final - E initial = E products – E reactants
Energy is a state function , a property dependent only on the current state of the system, not the path the system took to reach that state.
A change in the energy of the system is always accompanied by an opposite change in the energy of the surroundings. E final < E initial ΔE < 0 E final > E initial ΔE > 0
When energy is transferred from one object to another, it appears as work and/or as heat.
ΔE = q + w
- q = heat flowing into a system
- q = heat flowing out of a system
First Law of Thermodynamics
- the total energy of the universe is constant
- also called the Law of Conservation of Energy
- Energy is conserved: the energy of the system plus the energy of the surroundings remains constant ΔE universe = ΔE system + ΔE surroundings
Calorimetry Problem
The addition of hydrochloric acid to a silver nitrate solution precipitates silver chloride according to the following reaction:
Ag NO 3 (aq) + HCl (aq) AgCl (s) + HNO 3 (aq)
When 50.0 mL of 0.100 M AgNO 3 is combined with 50.0 mL of 0.100 M HCl in a coffee-cup calorimeter, the temperature changes from 23.40°C to 24.21°C. Calculate ΔHrxn for the reaction as written. Use 1.00 g/mL as the density of the solution and C = 4.18 J g-1^ °C-1^ as the specific heat capacity.
First, let’s draw a picture about what’s happening:
AgCl
Second, let’s remember the basic idea of calorimetry: Σ q = 0 So, that means that no heat escapes the calorimeter. If we can label the parts of this experiment that either generate or absorb heat, we should be able to find the heat of the reaction indirectly.
qrxn + qsoln + qcalorimeter = 0
Since the calorimeter’s heat capacity is not given, we will assume that the heat absorbed by the calorimeter is negligible, so we will eliminate the qcalorimeter from the above equation: qrxn + qsoln = 0
So, qrxn = - qsoln
where qsoln = msolncsoln Δ Tsoln
Also, remember:
" H (^) rxn =
q (^) rxn moles (acid)
50.0mL HCl
50.0 mL AgNO 3
50.0mL HCl
Third, let’s label the numbers given in the problem using these variables: T i = 23.40°C T f = 24.21°C C soln = 4.18 J g-^1 °C-^1 d soln = 1.00 g/mL m soln = 50.0 mL HCl + 50.0 mL AgNO 3 =
100.0 mL x
1.00 g mL
= 100.0 g Δ Hrxn =? (kJ/mol acid)
Fourth, let’s plug in the numbers:
qsoln = 100.0g x 4.
J
g o C
x (24.21 - 23.40) o^ C = 338.58 J
qrxn = - qsoln = - 338.58 J
We still need the moles of acid:
moles (HCl) = 50.0mL HCl x 0.100 moles HCl 1000 mL HCl
= 0.00500 moles HCl
Plug qrxn and the moles of acid into the ΔHrxn equation:
" Hrxn =
-338.58 J
0. 00500 moles HCl
J
mole
kJ mole
