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MRE_Yr 12 Hess Law

Authored by Mark Ebden

Chemistry

12th Grade

Used 19+ times

MRE_Yr 12 Hess Law
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15 questions

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1.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Using the equations below:


C(s) + O2(g) → CO2(g) ∆H = –390 kJ

Mn(s) + O2(g) → MnO2(s) ∆H = –520 kJ


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


MnO2(s) + C(s) → Mn(s) + CO2(g)

910

130

-130

-910

2.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

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


C(s) +O2(g) CO2(g) ΔH = –x kJ mol–1

CO(g) + O2(g) CO2(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) + O2(g) CO(g)

x + y

-x - y

y - x

x - y

3.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

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

4.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

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.

5.

MULTIPLE SELECT QUESTION

3 mins • 1 pt

Which of the following statements are true for the reaction:


SO2(g) + 1/2O2(g) ↔ SO3(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.

6.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Media Image

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

7.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Standard conditions are defined as...

298K and 1.00 x 105 kPa

273K and 1.00 x 105 kPa

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