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ENTHALPY, PHASE CHANGE, HESS' LAW

Total questions: 40

Worksheet time: 1hrs 5mins

Name
Class
Date
1.

How much energy is needed to melt 9.01g of ice at -12°C and convert it to steam at 100°C ?

a)

953,000 J

b)

9.53 kJ

c)

9.53 J

d)

95,337 kJ

2.
The phase change from water vapor to liquid water is known as...
a)
evaporation
b)
precipitation
c)
condensation
d)
sublimation
3.
2. A change of state from a liquid to a solid is called....
a)
Melting
b)
Freezing
c)
Evaporation
4.
5. What is it called when a solid turns directly into a gas?
a)
Sublimation
b)
Condensation
c)
Liquid
5.
Boiling is ...
a)
liquid to gas
b)
gas to solid
c)
gas to liquid
d)
solid to liquid
6.
Freezing is...
a)
Solid to gas
b)
Liquid to solid
c)
Gas to solid
d)
Liquid to gas
7.
For the formula:
 Q= m c ∆T
The  units for specific heat are:
a)
g /J C
b)
°C/g J
c)
kJ/g
d)
J/g°C
8.
What is the specific heat of an unknown substance if 100.0 g of it at 200.0 °C reaches an equilibrium temperature of 27.1 °C when it comes in contact with a calorimeter of water.  The water weighs 75. g and had an initial temperature of 20.00 °C?  (Specific heat of water is 4.18 J/g°C) (show your work)
a)
0.111 J/g°C
b)
1.29 J/g°C
c)
0.129 J/g°C
d)
22225.85 J
9.
Calorimetry Question:
A piece of metal with a mass of 32.8 g is heated to 100.5°C and dropped into 138.2 g of water at 20.0°C.  The final temperature of the system is 30.2°C.  What is the specific heat capacity of the metal? (show your work)
a)
2.56 J/g°C
b)
0.391 J/g°C
c)
5.29 J/g°C
d)
3.50 J/g°C
10.

A metal cube at temperature of 70°C is immersed in water at temperature of 20°C. The final temperature of the mixture will be

a)

Between 20°C and 70°C

b)

More than 70°C

c)

Less than 20°C

d)

Same as the room temperature

11.

A metal cube with a mass of 55grams at temperature of 85°C is immersed in 150grams of water at temperature of 20°C. The metal and water then reach a final temperature of 23oC. What is the specific heat of the metal?

a)

0.55 J/goC

b)

2120 J/goC

c)

11.4 J/goC

d)

4.18 J/goC

e)

0.75 J/goC

12.

You have dropped your popsicle and are watching it become a LIQUID. Which answer BEST explains what PHASE CHANGE has occurred AND how the molecule MOVEMENT has changed?

a)

The PHASE CHANGE that has occurred is CONDENSATION and the molecule MOVEMENT has become FASTER.

b)

The PHASE CHANGE that has occurred is EVAPORATION and the molecule MOVEMENT has become SLOWER.

c)

The PHASE CHANGE that has occurred is MELTNG and the molecule MOVEMENT has become FASTER.

13.

In the melting process, when ice and water are both present, the temperature will:

a)

stay at 0ºC

b)

decrease slowly

c)

increase slowly

d)

first increase then decrease

14.
Between which points is the temperature of the substance remaining constant?
a)
A-B only. 
b)
A-B, C-D, E-F
c)
B-C only. 
d)
B-C, D-E
15.

Absolute Zero

a)

The temperature of 0 C and pressure of 1 atm. A standard set of conditions to make comparisons

b)

The measure of heat

c)

The theoretical temperature where all motion stops. Zero heat energy

d)

The flow of heat from hot to cold

16.

Match the definition below to the correct term.

The enthalpy change that takes place when one mole of a compound is formed from its elements in their standard states under standard conditions.

a)

The enthalpy of neutralisation

b)

The enthalpy of combustion

c)

The enthalpy of formation

d)

The enthalpy of reaction

17.

How do you calculate Enthalpy of a reaction?

a)

ΔH = ΔHproducts - ΔHreactants

b)

ΔT = q / mC

c)

ΔG = ΔH -TΔS

d)

E = mc2

18.
The SI unit of heat and energy is the __________.
a)
calorie
b)
heat
c)
joule
d)
watt
19.
C+ O2 --> CO2 + 60kJ, what is the value for ΔH for the reaction?
a)
+60
b)
-60
c)
there is no way to know
20.
Endothermic reactions feel
a)
warm
b)
cold
21.

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)

a)

910

b)

130

c)

-130

d)

-910

22.

Using the equations below


Cu(s) + 1/2O2(g) → CuO(s) ∆H = –156 kJ

2Cu(s) + O2(g) → Cu2O(s) ∆H = –170 kJ


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


2CuO(s) → Cu2O(s) + 1/2O2(g)

a)

142

b)

15

c)

-15

d)

-142

23.

Consider the following equations.


Mg(s) + O2(g) → MgO(s) ∆H = –602 kJ

H2(g) + O2(g) → H2O(g) ∆H = –242 kJ


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


MgO(s) + H2(g) → Mg(s) + H2O(g)

a)

-844

b)

-360

c)

+360

d)

+844

24.

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)

a)

x + y

b)

-x - y

c)

y - x

d)

x - y

25.

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


P4(s) + 3O2(g) → P4O6(s) ΔHf = –1600 kJ mol–1

P4(s) + 5O2(g) → P4O10(s) ΔHf = –3000 kJ mol–1


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


P4O6(s) + 2O2(g) → P4O10(s)

a)

+4600

b)

+1400

c)

–1400

d)

–4600

26.
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 
a)
-677 kJmol–1 
b)
+111 kJmol–1 
c)
-111 kJmol–1 
d)
+677 kJmol–1 
27.
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) 
a)
394 + (2 × 286) – 891 
b)
–394 – (2 × 286) + 891 
c)
394 + 286 – 891 
d)
–394 – 286 + 891 
28.
Given the following data: ΔH○f[FeO(s)] = –270kJmol–1
ΔH○f [Fe2O3(s)] = –820 kJ mol–1
Select the expression which gives the enthalpy change, in kJ mol–1, for the reaction:
2FeO(s) + 1⁄2O2(g) → Fe2O3(s) 
a)
(–820 × 1⁄2) + 270 = –140
b)
(+820 × 1⁄2) – 270 = +140
c)
–820 + (270 × 2) = –280 
d)
+820 – (270 × 2) = +280 
29.
The standard enthalpy changes of formation of iron(II) oxide, FeO(s), and aluminium oxide, Al2O3(s), are –266 kJ mol–1 and –1676 kJ mol–1 respectively.
What is the enthalpy change under standard conditions for the following reaction?
3FeO(s) + 2Al (s)   ->   3Fe(s) + Al2O3(s) 
a)
+878kJ 
b)
–878kJ 
c)
–1942kJ 
d)
–2474kJ 
30.
As someone is running on the track they begin to perspire.  If the runner is our system, are they endothermic or exothermic?
a)
Endothermic process
b)
Exothermic process
31.
a)
-296.1
b)
226
c)
-11
d)
255.95
32.
a)
226
b)
255
c)
233
d)
11.3
33.
a)
-233
b)
-11.3
c)
-805
d)
-226
34.
a)
256
b)
202
c)
233
d)
804
35.
a)
-804.6
b)
-202.3
c)
-296.1
d)
-233.0
36.

2H2O ---> 2H2+ O2

2HNO3 ---> N2O5 + H2O

N2 + 3O2 + H2 ---> 2HNO3


What is the final net equation?

(Clue: Total the chemical reactions above using algebra method)

a)

H2O + N2 + 2O2 --> N2O5 + H2 +H2O

b)

H2O + N2 + 2O2 --> N2O5 + H2

c)

H2O + N2 + 2O2 --> N2O5 + H2 +H2O

d)

H2O + 2O2 --> N2O5 + H2 +H2O

37.

From the following enthalpy changes,

XeF2 (s) → Xe (g) + F2 (g) ∆H° = +123 kJ

Xe (g) + 2F2 (g) → XeF4 (s) ∆H° = -262 kJ

calculate the value of ∆H° for the reaction

XeF2 (s) + F2 (g) → XeF4 (s).

(a)  

38.

2 NO --> N2 + O2 (ΔH = -180.5 kJ)

N2 + 2 O2 --> 2 NO2 (ΔH = + 66.36 kJ)


Calculate the entalphy for :

2 NO + O2 --> 2 NO2 (ΔH = ?)

State exothermic or endothermic process?

a)

-114.14 kJ, exothermic

b)

+114.14 kJ, endothermic

c)

246.9 kJ, endothermic

d)

-246.9 kJ, exothermic

39.

“if you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction”

a)

Hess’s law.

b)

Avogadro’s law

c)

Boyle’s law.

40.

a)

-853.9 kJ

b)

853.9 kJ

c)

2498.1 kJ

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