Hess Law

Using the equations below:

C_(s) + O_2(g) → CO_2(g) ∆H = –390 kJ

Mn_(s) + O_2(g) → MnO_2(s) ∆H = –520 kJ

what is ∆H (in kJ) for the following reaction?

MnO_2(s) + C_(s) → Mn_(s) + CO_2(g)

• 910

• 130

• -130

• -910

Tags

NGSS.HS-PS1-4

Using the equations below

Cu_(s) + ¹/₂O_2(g) → CuO_(s) ∆H = –156 kJ

2Cu_(s) + O_2(g) → Cu₂O_(s) ∆H = –170 kJ

what is the value of ∆H (in kJ) for the following reaction?

2CuO_(s) → Cu₂O_(s) + ¹/₂O_2(g)

• 142

• 15

• -15

• -142

Tags

NGSS.HS-PS1-4

NGSS.HS-PS3-1

Consider the following equations.

Mg_(s) + O_2(g) → MgO_(s) ∆H = –602 kJ

H_2(g) + O_2(g) → H₂O_(g) ∆H = –242 kJ

What is the ∆H value (in kJ) for the following reaction?

MgO_(s) + H_2(g) → Mg_(s) + H₂O_(g)

• -844

• -360

• +360

• +844

Tags

NGSS.HS-PS1-4

The following equations show the oxidation of carbon and carbon monoxide to carbon dioxide.

C_(s) +O_2(g) → CO_2(g) ΔH = –x kJ mol^–1

CO_(g) + O_2(g) → CO_2(g) ΔH = –y kJ mol^–1

What is the enthalpy change, in kJ mol^–1, for the oxidation of carbon to carbon monoxide?

C_(s) + O_2(g) → CO_(g)

• x + y

• -x - y

• y - x

• x - y

Tags

NGSS.HS-PS1-4

Hess' Law makes use of which principle to calculate the enthalpy change of a reaction?

• The law of conservation of energy

• The law of conservation of matter

• The law that you will always find a lost item in the last place you look for it

• Murphy's law

Tags

NGSS.HS-PS3-1

For Hess' Law to be used what must be the same for all of the reactions being studied?

• The initial conditions of pressure and temperature.

• The final conditions of pressure and temperature.

• The initial and the final conditions of pressure and temperature.

• The initial and the final conditions of pressure and temperature, and the number of moles of reactants.

Which of the following statements are true for the reaction:

SO_2(g) + ¹/₂O_2(g) ↔ SO_3(g) ΔH = –92 kJ mol^-1

Where ↔ indicates that the reaction can proceed in the forward and the reverse direction.

• The forward and reverse reaction both produce 92 kJ of energy.

• Oxidising 2 moles of SO₂ would produce twice as much energy.

• The reverse reaction has an enthalpy of +92 kJ mol^-1.

• Collecting the SO3 produced in the liquid state would not change the measured enthalpy.

Tags

NGSS.HS-PS1-4

NGSS.HS-PS3-4

In order to find the enthalpy of combustion of C₃H₈ how must the enthalpy changes be arranged?

• ΔH₃ = ΔH₁ + ΔH₂

• ΔH₂ = ΔH₃ - ΔH₁

• ΔH₁ = ΔH₂ - ΔH₃

• 0 = ΔH₁ + ΔH₂ + ΔH₃

Tags

NGSS.HS-PS1-4

NGSS.HS-PS3-1

The standard enthalpy change of formation values of two oxides of phosphorus are:

P_4(s) + 3O_2(g) → P₄O_6(s) ΔH_f = –1600 kJ mol^–1

P_4(s) + 5O_2(g) → P₄O_10(s) ΔH_f = –3000 kJ mol^–1

What is the enthalpy change, in kJ mol^–1, for the reaction below?

P₄O_6(s) + 2O_2(g) → P₄O_10(s)

• +4600

• +1400

• –1400

• –4600

Tags

NGSS.HS-PS1-4

NGSS.HS-PS3-1

Standard conditions are defined as...

• 298K and 1.00 x 10⁵ kPa

• 273K and 1.00 x 10⁵ kPa