Enthalpy of Reaction with Molar Values

How do you calculate Enthalpy of a reaction?

• ΔH = ΔH_products - ΔH_reactants

• ΔT = q / mC

• ΔG = ΔH -TΔS

• E = mc²

Tags

NGSS.HS-PS3-1

Define standard enthalpy of combustion.

• Heat released when one mole of substance is burnt completely in excess oxygen.

• Heat absorbed when one mole of substance is burnt completely in excess oxygen under standard state.

• Heat released when one mole of substance is burnt completely in excess oxygen under standard state.

• Heat change when one mole of substance is burnt partially in excess oxygen under standard state.

Tags

NGSS.HS-PS1-4

Which of the following has a ΔH_f^o value of 0?

• Br₂(g)

• N(g)

• CO(g)

• Ne(g)

Which of the equation below refers to the standard enthalpy of formation, ΔH_f^o?

• Na(g) ---> Na⁺(g) + e^- ΔH = -364 kJmol^-1

• C₂H₅OH(l) + 3O₂(g) ---> 2CO₂(g) + 3H₂O (l) ΔH = - 1286 kJmol^-1

• 2C(s) + 2H₂(g) ---> C₂H₄ (g) ΔH = - 52.3 kJmol^-1

• Na⁺(g) ---> Na⁺(aq) ΔH = - 364 kJmol^-1

Tags

NGSS.HS-PS1-4

NGSS.HS-PS3-4

When 1.0 mole of ZnO(s) decomposes,

ZnO(s) ---> Zn(s) + 1/2 O₂(g) , enthalpy change is +348 kJ/mol.

What does this tell you about the formation of ZnO (s)?

• the formation of ZnO (s) is endothermic

• the formation of ZnO (s) is exothermic

• the formation of ZnO (s) does not require energy

• the formation of ZnO (s) absorbs heat.

Tags

NGSS.HS-PS1-4

Define 'Enthalpy'

• a) Energy stored in the movement of molecules in a substance
• b) The opposite of temperature
• c) The temperature of a molecule
• d) Energy stored in the chemical bonds in a substance

Tags

NGSS.HS-PS1-4

If ΔH is positive, heat would be shown on the _____ side of the thermochemical equation.

• Reactant
• Product

Tags

NGSS.HS-PS3-1

Calculate the enthalpy of the following reaction:
Ca(OH)₂ -> CaO + H₂O
ΔHf in kJ/mol:  
      Ca(OH)₂  -983.2
        CaO         -634.9
        H₂O         -285.5

• 62.8 kJ
• -62.8 kJ
• -1840.6
• 1840.6

Tags

NGSS.HS-PS1-4

NGSS.HS-PS3-1

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

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=a

H₂(g) + ½O₂(g) -> H₂O(l) ∆H=b

C₄H₉OH(l) + 6O₂(g) -> 4CO₂(g) + 5H₂O(l) ∆H=c

What is the enthalpy change for the reaction shown below?

4C(g) + 5H₂(l) + ½O₂(g) -> C₄H₉OH(l)

• c – 4a – 5b

• 4a + 5b - c

• 2a + 10b - c

• 2a + 5b + c

Tags

NGSS.HS-PS1-2

NGSS.HS-PS1-4

NGSS.HS-PS3-1