Free Printable Chemistry Hess's Law

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Chemistry Hess's Law

Total questions: 25

Worksheet time: 22mins

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The standard enthalpy change for the combustion of graphite is −393.5 kJ/mol and that of diamond is −395.4 kJ/mol.

What is the enthalpy change for the reaction below, in kJ/mol?

C _{(s, graphite)} → C _{(s, diamond)}

−1.9

+1.9

−788.9

+788.9

Calculate the change in Enthalpy of the combustion reaction of liquid Ethanol (C2H5OH). When Ethanol has an enthalpy of -286 kJ/ mol, water is liquid and oxygen and carbon dioxide are in gas state.

-286 kJ/ mol

1366 kJ/mol

-2732 kJ/ mol

-1366 kJ/mol

The standard enthalpy changes of formation of carbon dioxide and of methanoic acid are −394 kJ/mol and −409 kJ/mol respectively. Calculate the enthalpy change for the reaction

H₂ _(g) + CO₂ _(g) → HCOOH _(l)

−803 kJ/mol

−15 kJ/mol

+803 kJ/mol

+15 kJ/mol

How many joules of heat are needed to raise the temperature of 10.0 g of aluminum from 22°C to 55°C, if the specific heat of aluminum is 0.90 J/gx°C?

297 Joules

0.003 Joules

297 J/gx°C

0.003 J/gx°C

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

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

The standard enthalpy changes of combustion of carbon, hydrogen and methane are shown in the table.
Which one of the following expressions gives the correct value for the standard enthalpy change of formation of methane in kJ mol^–1?
C(s) + 2H₂(g) → CH₄(g)

394 + (2 × 286) – 891

–394 – (2 × 286) + 891

394 + 286 – 891

–394 – 286 + 891

Given the following data: ΔH^○_f[FeO(s)] = –270kJmol^–1
ΔH^○_f [Fe₂O₃(s)] = –820 kJ mol^–1Select 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

C₂H₄(g) + H₂(g) -> C₂H₆(g) ∆H°=-137 kJ mol^-1
Which statement about this information is correct?

The total energy of the bonds broken in the reactants is greater
than the total energy of the bonds
formed in the product

The bonds broken and the bonds made are of the same strength

The total energy of the bonds broken in the reactants is less than the total energy of the bonds formed in the product

No conclusion can be made about the sums of the bond enthalpies in the product compared with the reactants

10.

Use the chart to answer the following question.
Which is correct about energy changes during bond breaking and bond formation?

11.

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

12.

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

13.

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

14.

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

15.

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.

16.

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.

17.

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

18.

Standard conditions are defined as...

298K and 1.00 x 10⁵ kPa

273K and 1.00 x 10⁵ kPa

19.

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

20.

2 CO + O₂ --> 2 CO₂ (ΔH=-283 kJ)

If the equation was flipped and multiplied by 2 what would it look like?

2 CO + O₂ --> 2 CO₂

4 CO + 2 O₂ --> 4 CO₂

4 CO₂ -->4 CO + 2 O₂

2 CO₂ -->2 CO + O₂

21.

4 P + 6 Cl₂ --> 4 PCl₃(ΔH=-640 kJ)

If the equation was divided by 2 what would the chemical equation look like?

4 P + 6 Cl₂ --> 4 PCl₃

2 P + 3 Cl₂ --> 2 PCl₃

4 PCl₃--> 4 P + 6 Cl₂

2 PCl₃ -->2 P + 3 Cl₂

22.

To calculate ΔH what would we have to do to equation (b)

N₂ + 2 O₂ --> 2 NO₂

Using the following 2 equations:

(a) N₂ + O₂ --> 2 NO (ΔH = 180 kJ)

(b) 2 NO₂ --> 2 NO + O₂ (ΔH = 112 kJ)

(b) flip it because O₂ is on the products (right)

(b) keep it as is, because O₂ is on the reactants (left)

(b) multiply by 2 because O₂ have a 2 coefficient in front

23.

To calculate ΔH what would we have to do to equation (a)

N₂ + 2 O₂ --> 2 NO₂

Using the following 2 equations:

(a) N₂ + O₂ --> 2 NO (ΔH = 180 kJ)

(b) 2 NO₂ --> 2 NO + O₂ (ΔH = 112 kJ)

(A) flip it because O2 and N2 are in the products (right)

(A) keep it as is, because O2 and N2 are in the reactants (left)

(A) multiply by 2 because N2 and O2 have a 2 coefficient in front

24.

To calculate the ΔH of the final equation what must be done to (b)

2NOCl ---> 2 NO + Cl₂

Using the following 2 equations:

(a) 2 NO --> N₂ + O₂ (ΔH = -180.6 kJ)

(b) N₂ + O₂ + Cl₂ --> 2 NOCl (ΔH = 103.4 kJ)

(b) as is because Cl is on the reactant side (left) on the final

(a) flipped because Cl is on the product side (right) in the final

(a) multiplied by 2 because Cl has a coefficient of 2 in front of NO

25.

What would ΔH be if (a) was kept as is and (b) was flipped and multiplied by 2

Pb + PbO₂ + 2 H₂SO₄ --> 2 H₂O + 2 PbSO₄

Using the following 2 equations:

(a) Pb + PbO₂ + 2 SO₃ --> 2 PbSO₄ (ΔH = -775 kJ)

(b) SO₃ + H₂O --> H₂SO₄ (ΔH = -113 kJ)

-549 kJ

-662 kJ

1001 kJ

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Chemistry Hess's Law

Total questions: 25

Worksheet time: 22mins

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