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Increasing energy with temp?PowerPoint Presentation

Increasing energy with temp?

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Increasing energy with temp?

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- The added energy in a substance that occurs as temperature increases is stored in modes of motion in the substance
- For any molecule – modes are vibration, translation, and rotation
- Solid bond vibrations
- Gases translation
- Liquid water – complex function…

- When heat is added to a phase it’s temperature increases (No, really…)
- Not all materials behave the same though!
- dq=CVdT where CV is a constant (heat capacity for a particular material)
- Or at constant P: dq=CpdT
- Recall that dqp=dH then: dH=CpdT
- Relationship between CV and Cp:

Where a and b are coefficients of isobaric thermal expansion and isothermal compression, respectively

- HOWEVER C isn’t really constant….
- C also varies with temperature, so to really describe enthalpy of formation at any temperature, we need to define C as a function of temperature
- Maier-Kelley empirical determination:
- Cp=a+(bx10-3)T+(cx10-6)T2
- Where this is a fit to experimental data and a, b, and c are from the fit line (non-linear)

- Enthalpies, H, are found by calorimetry
- Enthalpies of formation are heats associated with formation of any molecule/mineral from it’s constituent elements

- Measurement of heat flow (through temperature) associated with a reaction
- Because dH = q / dT, measuring Temperature change at constant P yields enthalpy

Problem

When 50.mL of 1.0M HCl and 50.mL of 1.0M NaOH are mixed in a calorimeter, the temperature of the resultant solution increases from 21.0oC to 27.5oC. Calculate the enthalpy change per mole of HCl for the reaction carried out at constant pressure, assuming that the calorimeter absorbs only a negligible quantity of heat, the total volume of the solution is 100. mL, the density of the solution is 1.0g/mL and its specific heat is 4.18 J/g-K.

qrxn = - (cs solution J/g-K) (mass of solution g) (DT K)

= - (4.18 J/g-K) [(1.0g/mL)(100 mL)] (6.5 K)

= - 2700 J or 2.7 kJ

DH = 2.7 kJ

Enthalpy change per mole of HCl = (-2.7 kJ)/(0.050 mol)

= - 54 kJ/mol

Hess’s Law

Known values of DH for reactions can be used to determine DH’s for other reactions.

DH is a state function, and hence depends only on the amount of matter undergoing a change and on the initial state of the reactants and final state of the products.

If a reaction can be carried out in a single step or multiple steps, the DH of the reaction will be the same regardless of the details of the process (single vs multi- step).

CH4(g) + O2(g) --> CO2(g) + 2H2O(l) DH = -890 kJ

Net equation

CH4(g) + O2(g) --> CO2(g) + 2H2O(l) DH = -890 kJ

If the same reaction was carried out in two steps:

CH4(g) + O2(g) --> CO2(g) + 2H2O(g) DH = -802 kJ

2H2O(g) --> 2H2O(l) DH = -88 kJ

Hess’s law : if a reaction is carried out in a series of steps, DH for the reaction will be equal to the sum of the enthalpy change for the individual steps.