thermodymanics. thermodynamics is a branch of science that focuses on energy changes that accompany...

Thermodymanics

Thermodynamics is a branch of science that focuses on energy changes that accompany chemical and physical changes.

Objective: To calculate heat capacity.

Heat (q): The energy transferred between objects that are at different temperatures.

Unit: joules (J)

Molar Heat Capacity: Energy (heat) needed to increase the temperature of 1 mol of substance by 1 K.

q = nC∆T

q = heatn = # of molesC = molar heat capacity∆T = change in temperature

The molar heat capacity of water is larger than the molar heat capacity of land. This means that water does not heat up as easily as land does. As a result, oceans can help keep coastal areas cool during the summer.

The filling of a fruit pie has a larger heat capacity than the crust. This means that fruit filling will retain heat better and the crust will cool much quicker. As a result, eating the fruit filling can cause burns (even though it may appear that the pie is cool).

Determine the energy (heat) needed to increase the temperature of 10.0 mol of Hg by 7.5 K. The value of C for Hg is 27.8 J/K۰mol.

Determine the energy (heat) needed to increase the temperature of 10.0 mol of Hg by 7.5 K. The value of C for Hg is 27.8 J/K۰mol.

q = ?n = 10.0 molC = 27.8 J/K۰mol∆T = 7.5 K

q = nC∆Tq = (10.0 mol)(27.8 J/K۰mol)(7.5 K)

q= 2.1 x 103 J

The molar heat capacity of tungsten is 24.2 J/K ۰mol. Calculate the energy as heat needed to increase the temperature of 0.404 mol of W by 10.0 K.

The molar heat capacity of tungsten is 24.2 J/K ۰mol. Calculate the energy as heat needed to increase the temperature of 0.404 mol of W by 10.0 K.

q = ?n = 0.404 molC = 24.2 J/K۰mol∆T = 10.0 K

q = nC∆Tq = (0.404 mol)(24.2 J/K۰mol)(10.0 K)

q= 97.8J

Suppose a sample of NaCl increased in temperature by 2.5 K when the sample absorbed 1.7 x 102 J energy (heat). Calculate the number of moles of NaCl if the molar heat capacity is 50.5 J/K۰mol.

Suppose a sample of NaCl increased in temperature by 2.5 K when the sample absorbed 1.7 x 102 J energy (heat). Calculate the number of moles of NaCl if the molar heat capacity is 50.5 J/K۰mol.

q = 1.7 x 102 J n = ?C = 50.5 J/K۰mol ∆T = 2.5 K

q = nC∆T1.7 x 102 J = n(50.5 J/K۰mol)(2.5 K)

n= 1.3 mol

Calculate the energy as heat needed to increase the temperature of 0.80 mol of nitrogen, N2, by 9.5 K. The molar heat capacity of nitrogen is 29.1 J/K۰mol.

Calculate the energy as heat needed to increase the temperature of 0.80 mol of nitrogen, N2, by 9.5 K. The molar heat capacity of nitrogen is 29.1 J/K۰mol.

q = ?n = 0.80 molC = 29.1 J/K۰mol ∆T = 9.5 K

q = nC∆Tq = (0.80 mol)(29.1 J/K۰mol)(9.5 K)

q= 2.2 x 102 J

A 0.07 mol sample of octane, C8H18, absorbed 3.5 x 103 J of energy. Calculate the temperature increase of octane if the molar heat capacity of octane is 254.0 J/K۰mol.

A 0.07 mol sample of octane, C8H18, absorbed 3.5 x 103 J of energy. Calculate the temperature increase of octane if the molar heat capacity of octane is 254.0 J/K۰mol.

q = 3.5 x 103 J n = 0.07 molC = 254.0 J/K۰mol ∆T = ?

q = nC∆T3.5 x 103 J = (0.07 mol)(254.0 J/K۰mol) ∆T

∆T = 2.0 x 102 K

Objective: To calculate the change in enthalpy.

The total energy of a system is impossible to measure.

However, we can measure the change in enthalpy of a system.

Enthalpy: The total energy content of a sample.

H = Enthalpy ∆H = Change in Enthalpy

∆H can be measured with a calorimeter.

A calorimeter is used to measure the heat absorbed or released in a chemical or physical change.

Enthalpy changes can be used to determine if a process is endothermic or exothermic.

Exothermic Reaction: Negative Enthalpy Change

Endothermic Reaction: Positive Enthalpy Changes

The standard enthalpy of formation ( ) is the enthalpy change in forming 1 mol of a substance from elements in their standard states.

Note: The value for the standard enthalpy of formation for an element is 0.

The values for the standard enthalpies of formation can be found using a table.

0fH

ΔHreaction = ΔH products - ΔHreactants

Step 1: Determine for each compound using enthalpy table.

Step 2: Multiply by the coefficients from the balanced equation (# of moles).

Step 3: Set up ΔH equation. Step 4: calculate.

0fH


Calculate ΔH for the following reaction and determine if the reaction is exothermic or endothermic.

SO2(g) + NO2(g) SO3(g) + NO(g)


SO2(g) + NO2(g) SO3(g) + NO(g)

-296.8(1) 33.1(1) -395.8(1) 90.3(1)

kJ/molkJ/mol kJ/molkJ/mol

ΔH= [(-395.8 kJ/mol)(1) + (90.3 kJ/mol)(1)] – [(296.8

kJ/mol)(1) + (33.1 kJ/mol)(1)]

Δ H = -305.5 kJ – 329.9 kJ

Δ H = -635.4 kJ The reaction is exothermic.

Calculate the enthalpy change for the following reaction:

2C2H6(g) + 7O2(g) 4 CO2(g) + 6H2O(g)

Calculate ΔH for the following reaction:CaO(s) + H2O(l) Ca(OH)2(s)

Calculate the enthalpy change for the combustion of methane gas.

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

Objective: To calculate the change in entropy.

A reaction is more likely to occur if enthalpy (ΔH) is negative.

However, some endothermic reactions can occur easily. Why? Entropy!

Entropy (ΔS): A measure of the randomness or disorder of a system.

A process if more likely to occur if there is an increase in entropy (or if ΔS is positive).

Entropy is increased by the following factors:Diffusion (process of dispersion)Dilution of a solutionDecreasing the pressure of a gas Increasing temperatureThe number of moles of product is greater

than the number of moles of reactant Increasing the total number of particles in a

systemWhen a reaction produces more gas

particles (opposed to liquid or solids)

ΔSreaction = ΔSproducts - ΔSreactants

Find the change in entropy for the following reaction:

2Na (s) + 2HCl (g) 2NaCl (s) + H2(g)

Find the change in entropy for the following reaction:

2Na (s) + 2H2O (l) 2NaOH (s) + H2(g)

Objectives:(1)To calculate the change in Gibbs

energy.(2)To determine if a reaction is

spontaneous or nonspontaneous.

Gibbs Energy: the energy in a system that is available to do useful work.

Gibbs Energy is also called Free Energy.

ΔG = Change in Gibbs Energy

ΔGreaction = ΔGproducts - ΔGreactants

Calculate ΔG for the following water-gas reaction:

C(s) + H2O (g) CO (g) + H2 (g)

Calculate the Gibbs energy change that accompanies the following reaction:

C(s) + O2 (g) CO2 (g)

Calculate the Gibbs energy change that accompanies the following reaction:

CaCO3 (s) CaO (s) + CO2 (g)

If ΔG is negative, the forward reaction is spontaneous.

If ΔG is 0, the system is at equilibrium.

If ΔG is positive, the forward reaction is nonspontaneous. NOTE: In this case, the reaction is spontaneous in the

reverse direction.

ΔG = ΔH - TΔS

Step 1: Organize the information.Step 2: Change the units.Step 3: Step up ΔG equation.Step 4: Calculate.

Given the changes in enthalpy and entropy are -139 kJ and 277 J/K respectively for a reaction at 25⁰C, calculate the change in Gibbs energy.


ΔH = -139 kJΔS = 277 J/K / (1000 J/kJ) = 0.277 kJ/KT = 25 ⁰C + 273 = 298 KΔG = ?


ΔH = -139 kJΔS = 277 J/K / (1000 J/kJ) = 0.277 kJ/KT = 25 ⁰C + 273 = 298 KΔG = ?

ΔG = ΔH - T ΔSΔG = (-139 kJ) – (298K)(0.277 kJ/K)ΔG = (-139 kJ) – (82.546 kJ)ΔG = -221.55 kJ The reaction is spontaneous.

A reaction has a ΔH of -76 kJ and a ΔS of -117 J/K. Calculate the ΔG at 298 K. Is the reaction spontaneous?

A reaction has a ΔH of 11 kJ and a ΔS of 49 J/K. Calculate ΔG at 298 K. Is the reaction spontaneous?

Objective: To calculate the melting and boiling point of a substance.

Tmp: melting point temperature Tbp: boiling point temperature ΔHfus: molar enthalpy of fusion ΔSfus: molar entropy of fusion ΔHvap: molar enthalpy of vaporization ΔSvap: molar entropy of vaporization

fus

fusmp S

HT

vap

vapbp S

HT

Step 1: Organize the information. Step 2: Change the units. Step 3: Step up the equation. Step 4: Calculate.

The enthalpy of fusion of Hg is 2.295 kJ/mol, and the molar entropy of fusion is 9.79 J/mol·K. The enthalpy of vaporization at the boiling point if 59.2 kJ/mol, and the molar entropy of vaporization is 93.8 J/mol·K. Calculate the melting point and boiling point of Hg.


Tmp = ?

ΔHfus = 2.295 kJ/mol

ΔSfus = 9.79 J/mol·K = 0.00979 kJ/mol·K


Tmp = ?



Tmp = (2.295 kJ/mol) / (0.00979 kJ/mol·K)

Tmp = 234 K


Tmp = ?



Tmp = (2.295 kJ/mol) / (0.00979 kJ/mol·K)

Tmp = 234 K

Tbp = ?

ΔHvap = 59.2 kJ/mol

ΔSvap = 93.8 J/mol·K = 0.0938 kJ/mol·K


Tmp = ?



Tmp = (2.295 kJ/mol) / (0.00979 kJ/mol·K)

Tmp = 234 K

Tbp = ?

ΔHvap = 59.2 kJ/mol

ΔSvap = 93.8 J/mol·K = 0.0938 kJ/mol·K

Tbp = (59.2 kJ/mol) / (0.0938 kJ/mol·K )

Tbp = 631 K

For ethanol, the molar enthalpy of fusion is 4.931 kJ/mol and the molar entropy of fusion is 31.6 J/mol·K. The molar enthalpy of vaporization at the boiling point is 42.32 kJ/mol and the molar entropy of vaporization is 109.9 J/mol·K. Calculate the melting and boiling points for ethanol.

For sulfur dioxide, the molar enthalpy of fusion is 8.62 kJ/mol and the molar entropy of fusion is 43.1 J/mol·K. The enthalpy of vaporization at the boiling point is 24.9 kJ/mol and the molar entropy of vaporization is 94.5 J/mol·K. Calculate the melting and boiling points for sulfur dioxide.

For ammonia, ΔHfus is 5.66 kJ/mol and Δsfus is 29.0 J/mol·K. ΔHvap is 23.33 kJ/mol and ΔSvap is 97.2 J/mol·K. Calculate the melting and boiling points for ammonia.

thermodymanics. thermodynamics is a branch of science that focuses on energy changes that accompany...

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