Catalyst

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Catalyst Given the following equations and H o values, determine the heat of reaction (kJ) at 298 K for the reaction: B 2 H 6 (g) + 6 Cl 2 (g) 2 BCl 3 (g) + 6 HCl(g) BCl 3 (g) + 3 H 2 O(l) H 3 BO 3 (g) + 3 HCl(g) ΔH = -112.5 kJ B 2 H 6 (g) + 6 H 2 O(l) 2 H 3 BO 3 (s) + 6 H 2 (g ΔH = -493.4 kJ 1/2 H 2 (g) + 1/2 Cl 2 (g) HCl(g) Δ H = -92.3 kJ

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Catalyst. Given the following equations and H o values, determine the heat of reaction (kJ) at 298 K for the reaction: B 2 H 6 (g) + 6 Cl 2 (g)  2 BCl 3 (g) + 6 HCl(g) BCl 3 (g) + 3 H 2 O(l)  H 3 BO 3 (g) + 3 HCl(g) ΔH = -112.5 kJ - PowerPoint PPT Presentation

Transcript of Catalyst

Page 1: Catalyst

CatalystGiven the following equations and Ho values,

determine the heat of reaction (kJ) at 298 K for the reaction:

B2H6(g) + 6 Cl2(g) 2 BCl3(g) + 6 HCl(g) BCl3(g) + 3 H2O(l) H3BO3(g) + 3 HCl(g) ΔH = -

112.5 kJB2H6(g) + 6 H2O(l) 2 H3BO3(s) + 6 H2(g ΔH = -

493.4 kJ1/2 H2(g) + 1/2 Cl2(g) HCl(g) Δ H = -92.3

kJ

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Lecture 5.3 – Enthalpies of Formation and Bond Enthalpies

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Today’s Learning TargetsLT 5.6 – For a given chemical bond, I can

characterize and calculate its bond enthalpy. Furthermore, I can apply this value to the type and length of bond that the molecule contains.

LT 5.8 – For a phase change, I can characterize the enthalpy of formation and what this value indicates about a given phase change.

LT 5.9 – I can apply Hess’s Law to a chemical reaction and calculate the enthalpy for a combination of chemical reactions.

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Enthalpies of Formation (ΔHf)The enthalpy change for creating a compound

from its constituent elements is known ΔH of formation (ΔHf)

2 C (s)+ H2 (g) C2H2 ΔHf =226.7 kJ/molThe value of ΔHf depends on temperature,

pressure, and state of products/reactants

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Standard EnthalpiesIn order to compare ΔHf of different reactions,

we define standard state of 1 atm and 298 KDon’t confuse with STP! (1 atm and 273 K)

Standard Enthalpies (ΔHo) are ΔH values calculated under these standard conditions

ΔHfo is the change in enthalpy to produce one

mole of the compound from the constituent elements

All elements that are stable at room temperature have a ΔHf

o of 0 kJ/mol

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Class ExampleWrite the equation corresponding to the

standard enthalpy of formation of liquid carbon tetrachloride (CCl4).

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Table TalkWrite the equation corresponding to the

standard enthalpy of formation of gaseous propane (C3H8).

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Using ΔHf to calculate ΔHrxnWe can use known ΔHf

o and Hess’s Law to calculate the ΔHrxn

o for a given reaction of interest.For the combustion of propane gas, C3H8 (g), to

CO2 (g) and H2O (l) under standard conditions:C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4H2O (l) ΔHrxn

o = ?You know that:3 C (s) + 4H2 (g) C3H8 (g) ΔHf

o = -103.85 kJ/mol C (s) + O2 (g) CO2 (g) ΔHf

o = -393.5 kJ/mol H2 (g) + ½ O2 (g) H2O (l) ΔHf

o = -285.8 kJ/mol

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Shortcut for Using ΔHf to calculate ΔHrxnWe can quickly calculate the ΔHrxn by using the

equation:

Note that n and m represent the coefficient from the balanced chemical reaction.

ΔH rxno = ΣnΔH f

o(products) − ΣmΔH fo(reactants)

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Class ExampleCalculate the standard enthalpy change for

the combustion of 1 mole of benzene (C6H6) to CO2 (g) and H2O (l).

Note:ΔHf

o (CO2) = -393.5 kJ/molΔHf

o (C6H6) = 49.0 kJ/molΔHf

o (H2O) = -285.8 kJ/mol

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Table TalkCalculate the standard enthalpy change for the

combustion of 1 mole of ethanol (C2H5OH) to CO2 (g) and H2O (l).

Note:ΔHf

o (CO2) = -393.5 kJ/molΔHf

o (C2H5OH) = -277.7 kJ/molΔHf

o (H2O) = -285.8 kJ/mol

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Bond Enthalpies and ΔHrxnWe can use bond enthalpies to estimate the

ΔHrxn where certain bonds are broken and new bonds are formed.

Allows for a quick determination of exo- or endothermicNOTE – This is an estimate

This is an application of Hess’s LawAssume all bonds are broken and then all

bonds reform:

ΔHrxn = Σ(enthalpies of bonds broken) - Σ(enthalpies of bonds formed)

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Class ExampleYou run the reaction:

2 C2H6 (g) + 7 O2 (g) 4 CO2 (g) + 6 H2OUsing the following ΔH values, calculate ΔHrxn for

the reaction and determine if it is endothermic or exothermic:

ΔHC – H = 413 kJ/molΔHC – C = 348 kJ/molΔHO=O = 495 kJ/molΔHC =O = 799 kJ/molΔHH– O = 463 kJ/mol

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Table TalkYou run the reaction:

2 N2H4 (g) N2 (g) + 2 H2Using the following ΔH values, calculate ΔHrxn

for the reaction and determine if it is endothermic or exothermic:

ΔHN – H = 391 kJ/molΔHN – N = 163 kJ/molΔHN2 = 941 kJ/molΔHH-H = 436 kJ/mol

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Problem Set 5.3This is a sampling of free response type

questions that have been observed on past AP exams

This will be due next Monday/Tuesday!

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Rate YourselfWe have finished almost all of chapter 5!Rate yourself 1 – 4 on LTs 5.1 to 5.9.It is your responsibility to study any areas that

are a 2 or lower over the long weekend

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Happy Thanksgiving!

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Closing TimeClosing TimeRead 5.7, 8.8Do book problems: