Can you explain step by step...It\'s kind of confusing, thanks! Consider the rea

Question

Can you explain step by step...It's kind of confusing, thanks!

Consider the reaction:

CO (g) + H2O (g)H2 (g) + CO2 (g)

A. Use average bond dissociation energies to calculate Hrxn for the reaction. SHOW ALL WORK.

CO: 1072 kJ/mol O–H: 464 kJ/mol H–H: 436 kJ/mol C=O: 799 kJ/mol

B. Use standard enthalpies of formation (H°f) to calculate H°rxn for the reaction. SHOW ALL WORK. (Hint: You may get a different value of H°rxn as compared to part A.)

H°f (CO) = – 110.5 kJ/mol H°f (H2O) = – 241.8 kJ/mol H°f (CO2) = – 393.5 kJ/mol

C. Based on the H values from part A and B, is the given reaction exothermic or endothermic?

Explanation / Answer

A)

CO = 1072 kJ/mol, O-H = 464 kJ/mol, H-H = 436 kJ/mol, C=O = 799 kJ/mol,

let's break these down to atoms
CO (g) + H2O (g) ---> C + 2H + 2O
put in the energies needed to break bonds
CO (g) + 1072 + H2O (g) + 2(464) ---> C + 2H + 2O
there were 1 CO bonds and 2 O-H bonds
delta(H) = +2000 kJ (added energy)
now let's recombine the atoms to get the products
C + 2H + 2O ---> CO2 + 2(799) + H2 + (436)
delta(H) = -2034 kJ (energy releases add the two delta(H) terms
Hrxn = +2000 + (-2034) = -34 kJ

B)  Hrxn = p*H(products) - rH(reactants)

where p and r are the corresponding stoichiometric coefficieints from the balanced reaction.

CO (g) + H2O (g) ---> CO2 (g) + H2 (g)

H°f (CO) = – 110.5 kJ/mol H°f (H2O) = – 241.8 kJ/mol H°f (CO2) = – 393.5 kJ/mol

Because H2 is an element in its natural state, its Hf is 0.

Hrxn = (H(CO2(g)) + H(H2(g))) - (H(CO(g)) + H(H2O(g))
Hrxn = ((393.5) + (0)) - (110) +(-241.8)

Hrxn = ((393.5) + (351.8)

Hrxn = - 41.7  kJ/mol

C) Exothermic = Energy released; Endothermic = Energy kept in does that mean Exothermic = (-delta H)
and Endothermic = (+delta H)

Part A and B, is the given reaction H = -ve values (-34 kJ and   - 41.7  kJ/mol)

so above reaction is Exothermic.

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