Question

Using the following thermal chemical data (use Hess’s law) 2Fe(s) + 6HF(g) —> 2FeF3(s) + 3H2(g)...

Using the following thermal chemical data (use Hess’s law)

2Fe(s) + 6HF(g) —> 2FeF3(s) + 3H2(g)   
?rH•= -1787.4 kJ/mol
2Fe(s) + 6HCl(g) —> 2FeCl3(s) + 3H2(g)
?rH•= -1457.0 kJ/mol

calculate?rH• for the following reaction:
FeCl3(s) + 3HF(g) —> FeF3(s) + 3HCl(g)

2. when 19.86g NaOH is dissolved in 125 mL of water in the coffee-cup calorimeter, the temperature rises from 23•C to 65•C. what is the enthalpy change per mole of the hydroxide dissolved in the water? Assume that the solution has a specific heat capacity of 4.18 J/gK

3. A 2.5 g sample but of octane (C8H18) is burned in a bomb calorimeter containing 0.75 kg of water at an initial temperature of 24•C. After the reaction, the final temperature of the water is 32•C. the heat capacity of the calorimeter is 923 J/•C. the specific heat of water is 4.184 J/g•C. calculate the heat of combustion of octane in kJ/mol.

Homework Answers

Answer #1

Ans 1

The first reaction is

2Fe(s) + 6HF(g) —> 2FeF3(s) + 3H2(g)   

multiply by 0.5

Fe(s) + 3HF(g) —> FeF3(s) + 1.5H2(g)

H1 = -1787.4*0.5 = - 893.7 kJ/mol

The second reaction is

2Fe(s) + 6HCl(g) —> 2FeCl3(s) + 3H2(g)

Reverse the reaction and multiply by 0.5

FeCl3(s) + 1.5H2(g) = Fe(s) + 3HCl(g)

H2 = 1457.0*0.5 = 728.5 kJ/mol

Add the two reactions

Fe(s) + 3HF(g) + FeCl3(s) + 1.5H2(g) —> FeF3(s) + 1.5H2(g) + Fe(s) + 3HCl(g)

FeCl3(s) + 3HF(g) —> FeF3(s) + 3HCl(g)

H = H1 + H2

= - 893.7 + 728.5

= - 165.2 kJ/mol

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