Hess's law and bond enthalpies

The enthalpy change for the reaction, ∆H_r , is equal to

• ∆H₁ + ∆H₂

• ∆H₁ - ∆H₂

• -∆H₁ - ∆H₂

• -∆H₁ + ∆H₂

Tags

NGSS.HS-PS1-4

The enthalpy change for the reaction

C(s, graphite) + 1⁄2O₂(g) --> CO(g) cannot be measured directly since some carbon dioxide is always formed in the reaction.

It can be calculated using Hess’s Law and the enthalpy changes of combustion of graphite and of carbon monoxide.

C(s, graphite) + O₂(g) --> CO₂ ΔH=-394 kJmol^–1

CO(g) + 1⁄2O₂(g) --> CO₂ ΔH=-283 kJmol^–1

The enthalpy change for the reaction of graphite with oxygen to give carbon monoxide is

• -677 kJmol^–1

• +111 kJmol^–1

• -111 kJmol^–1

• +677 kJmol^–1

What is the relationship between enthalpies p, q, r and s?

S(s) + H₂(g) -> H₂S(g) ∆H=p

H₂(g) + ½O₂(g) -> H₂O(l) ∆H=q

S(s) + O₂(g) -> SO₂(g) ∆H=r

H₂S(g) + 1½O₂(g) -> H₂O(l) + SO₂(g) ∆H=s

• p = q + r – s

• p = s – q – r

• p = q – r – s

• p = s + r – q

Tags

NGSS.HS-PS1-4

4. The enthalpies of combustion of C(s), H2(g) and C4H9OH(l) (in kJmol^-1) are as follows

C(s) + O₂(g) -> CO₂(g) ∆H=-393kJ/mol

H₂(g) + ½O₂(g) -> H₂O(l) ∆H=-286kJ/mol

C₄H₉OH(l) + 6O₂(g) -> 4CO₂(g) + 5H₂O(l) ∆H=-1371kJ/mol

What is the enthalpy change for the reaction shown below?

4C(g) + 5H₂(l) + ½O₂(g) -> C₄H₉OH(l)

• +845kJ

• -2275 kJ

• -1631kJ

• -845

Tags

NGSS.HS-PS1-4

The standard enthalpy changes of formation of iron(II) oxide, FeO(s), and aluminium oxide, Al₂O₃(s), are –266 kJ mol^–1 and –1676 kJ mol^–1 respectively.

What is the enthalpy change under standard conditions for the following reaction?

3FeO(s) + 2Al (s)  -> 3Fe(s) + Al₂O₃(s)

• +878kJ

• –878kJ

• –1942kJ

• –2474kJ

The average bond enthalpies for O-O and O=O are 146kJ/mol and 496kJ/mol respectively. Calculate the enthalpy change for the reaction shown.

• -102kJ

• +102kJ

• +146kJ

• -248 kJ

Tags

NGSS.HS-PS1-4

NGSS.HS-PS3-1

The average bond enthalpy for the C-H bond is 412kJ/mol. which process has an enthalpy change closest to this value?

• 
• 
• 
• 

Given the following data: ΔH^○_f[FeO(s)] = –270kJmol^–1

ΔH^○_f [Fe₂O₃(s)] = –820 kJ mol^–1

Select the expression which gives the enthalpy change, in kJ mol^–1, for the reaction:

2FeO(s) + 1⁄2O₂(g) → Fe₂O₃(s)

• (–820 × 1⁄2) + 270 = –140

• (+820 × 1⁄2) – 270 = +140

• –820 + (270 × 2) = –280

• +820 – (270 × 2) = +280