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Hess’ Law

Total questions: 20

Worksheet time: 14mins

Name

Class

Date

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

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

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

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

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.

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

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

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

-142

10.

A reaction is performed in a beaker with a temperature probe recording the temperature changes of the reaction. If the temperature began at 15.0 degrees Celsius and ended at 27.5 degrees Celsius. If the reaction is our system, is the system endothermic or exothermic?

Exothermic

Endothermic

11.

In an endothermic reaction the system is releasing energy.

True

False

Sometimes

12.

A chemical reaction that requires energy is a

exothermic reaction

endothermic reaction

over reaction

redox reaction

13.

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

2a + 10b - c

4a + 5b - c

2a + 5b + c

14.

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) --> CO₂ΔH=-394 kJmol^–1
CO(g) + 1⁄2O₂(g) --> CO₂ΔH=-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

15.

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

16.

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

17.

What is the Delta H if you reverse the reaction.

86 kJ

- 86 kJ

68 kJ

-68 kJ

18.

How much energy is associated with the process of creating 4 mol NO2

1933 kJ

-1875 kJ

3866 kJ

-3750 kJ

19.

In rnx 2 we have -572 kJ instead of - 286kJ. Why?

We had to multiple the reaction by 2 to get 2 mole H2 in reactants.

We had to multiple the reaction 2 to get 2 moles H2O on the product side to cancel the 2 moles of H2O in rnx 3.

Both A and C

Neither A or C

20.

Why do we have 891 kJ in rnx 3 instead of the original -891 kJ?

need to reverse reaction to get 1 mole of CH4 on product side.

need to reverse reaction to get 2 mole of O2 on product side

need to cancel the 2 moles of O2 on the reactant side when you sum rnx 1 and 2

all of the above

100 %

Hess’ Law

Total questions: 20

Worksheet time: 14mins

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