10/02/2016

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Enthalpy of reaction

ΔH = Hfinal Hinitialso in chemical reactions ΔH = Hproducts Hreactants

ΔH is called the enthalpy of reaction,

Thermochemical equation is the balanced chemical reaction and the value of ΔH

provides relationship between amounts of chemicals and the enthalpy change. Things to note about enthalpy:

1. ΔH depends on amounts of reactants and products

2. ΔH forward = - ΔH reverse

3. ΔH depends on the physical state of reactants and products

written as ΔHrxn

2H2O2(ℓ) 2H2O(ℓ) + O2(g) ΔH = 196 kJ

H2O2(ℓ) H2O(ℓ) + ½O2(g) ΔH = ½[196 kJ]

2H2O2(ℓ) 2H2O(g) + O2(g) ΔH = 108 kJ

2H2O2(ℓ)

2H2O(ℓ) + O2(g)

ΔH = 196 kJ

ΔH = +196 kJexothermic

endothermic2H2O(ℓ) + O2(g) 2H2O2(ℓ) ΔH = +196 kJ

Question:

It is said that James Joule (1818 – 1889) was almost fanatical about measurement.

On his honeymoon he took a very sensitive thermometer and measured the

temperature of water at the top and at the bottom of a scenic waterfall.

You would expect the water at the bottom of the waterfall to be …

A) a little cooler than the water at the top

B) a little warmer than the water at the top

C) exactly the same temperature as the water at the top

D) harder to get to

E) Wetter

Not as simple as it might seem! Kinetic energy of falling water converted into heat

will warm the water. Evaporation as the liquid falls will cool it…

10/02/2016

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Question:

Which of these is NOT a state function?

a. internal energy

b. temperature

c. enthalpy

d. work

Question:

Which of these is NOT a state function?

a. internal energy

b. temperature

c. enthalpy

d. work

10/02/2016

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Specific heat can be determined experimentally

e)temperatur in (change x substance)of (grams

dtransferre heatof amountheat specific

An example: determining the specific heat of lead:

(a) Heat 150g lead to 100°C(b) Add the hot lead to 50g water at 22°C (c) Measure the final temperature

Rearrange to give: q = CS x mass x ΔT

q lost = Cs(Pb) x 150.0 g x (28.8 - 100.0)°C

q gained = Cs(H2O) x 50.0 g x (28.8 – 22.0)°Cheat lost = heat gained, hence calculate Cs(Pb)

Suppose Tfinal is 28.8 oC

Hess’s Law

ΔH is a state function, so does not depend upon how we get from reactants to products

We can use tabulated ΔH values to calculate the enthalpy of reactions

ΔH depends on amounts of reactants and products and their initial and final states

Example: N2(g) + 3H2(g) 2 NH3(g) Hrxn = ?

N2(g) + 3H2(g) N2H4(g) + H2(g) H = 95.4 kJ

N2H4(g) + H2(g) 2 NH3(g) ΔH = 187.6 kJ

step 1ΔH = 95.4 kJ

Hrxn = + 95.4 kJ -187.6 kJ = 92.2 kJ

step 2ΔH = -187.6 kJ

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Hess’s Law : The enthalpy change of an overall process is the sum of the

enthalpy changes of its individual steps

combine the individual reactions so their sum gives thedesired reaction

Procedure:

• Arrange reactions so (final) reactants appear onthe left and products appear on the right

• Any reaction that is reversed must have the sign of its ΔH changed

• All intermediates must occur on both the right and the left so they cancel

• A reaction can be multiplied by a coefficient, but ΔH for that reaction must be multiplied by the samecoefficient.

e.g. what is ΔH for ½N2(g) + O2(g) → NO2(g)given:N2(g) + O2(g) 2NO( g) ΔH = +180.50 kJNO2(g) NO(g) + ½ O2(g) ΔH = +57.07 kJ

NO(g) + ½ O2(g) → NO2(g) ΔH = 57.07 kJ-

N2(g) + O2(g) → 2NO( g) ΔH = +180.50 kJ

½N2(g) + ½O2(g) → NO( g) ΔH = (½) 180.50 kJ

½N2(g) + O2(g) → NO2(g) ΔH = -57.07 kJ + 90.25 kJ = 33.18 kJ

Hess’s Law

ΔH is a state function, so does not depend upon how we get from reactants to products

We can use tabulated ΔH values to calculate the enthalpy of reactions

ΔH depends on amounts and initial and final states of reactants and products

Example: N2(g) + 3H2(g) 2 NH3(g) Hrxn = ?

N2(g) + 2H2(g) N2H4(g) H = 95.4 kJ

NH3(g) ½ N2H4(g) + ½ H2(g) ΔH = + 93.8 kJ

Hrxn = -187.6 kJ + 95.4 kJ = 92.2 kJ

10/02/2016

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If X is a state function, then the change in X is given by (Xfinal …. Xinitial)

a. +

b. –

c. ×

d. ÷

If X is a state function, then the change in X is given by (Xfinal …. Xinitial)

a. +

b. –

c. ×

d. ÷

10/02/2016

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The standard enthalpy of formation of carbon in its graphite form is

_______ kilojoules per mole.

a. 100

b. 1000

c. 1

d. 0

The standard enthalpy of formation of carbon in its graphite form is

_______ kilojoules per mole.

a. 100

b. 1000

c. 1

d. 0

10/02/2016

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The standard enthalpy of formation of carbon in its diamond form is

+1.88 kJ/mole, which means that diamond is _______ graphite.

a. as stable as

b. more stable than

c. less stable than

d. an isotope of

The standard enthalpy of formation of carbon in its diamond form is

+1.88 kJ/mole, which means that diamond is _______ graphite.

a. as stable as

b. more stable than

c. less stable than

d. an isotope of

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2 H2 + O2 2 H2O

If the reaction above releases 483.6 kJ, then the standard enthalpy of

formation of H2O is

a. –483.6 kJ/mole.

b. +483.6 kJ/mole.

c. –241.8 kJ/mole.

d. +241.8 kJ/mole.

2 H2 + O2 2 H2O

If the reaction above releases 483.6 kJ, then the standard enthalpy of

formation of H2O is

a. –483.6 kJ/mole.

b. +483.6 kJ/mole.

c. –241.8 kJ/mole.

d. +241.8 kJ/mole.

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What effect does reversing a reaction have on the value of H?

A. No change.

B. Sign of H changes.

C. Value of H increases.

D. Value of H decreases.

……………………………………………………………………………….

What effect does reversing a reaction have on the value of H?

A. No change.

B. Sign of H changes.

C. Value of H increases.

D. Value of H decreases.

……………………………………………………………………………….

10/02/2016

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Ozone, O3(g), is a form of elemental oxygen produced by electric motors, during electrical storms and in the upper atmosphere. Is Hf for O3(g) necessarily zero?

A. Yes, because it is just another elemental form of oxygen.

B. No, because it is not the most stable form of the element oxygen at the given conditions.

C. Yes, because changing the subscripts of an elemental formula does not change standard enthalpy of formation.

D. No, because there is a temperature change when ozone is formed.

Ozone, O3(g), is a form of elemental oxygen produced by electric motors, during electrical storms and in the upper atmosphere. Is Hf for O3(g) necessarily zero?

A. Yes, because it is just another elemental form of oxygen.

B. No, because it is not the most stable form of the element oxygen at the given conditions.

C. Yes, because changing the subscripts of an elemental formula does not change standard enthalpy of formation.

D. No, because there is a temperature change when ozone is formed.

10/02/2016

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It’s now possible to buy cars that run on hydrogen gas as a fuel; what advantage(s) does hydrogen have as a fuel?

A. It is easily stored as a gas for a long time.

B. Its product of combustion is only H2O(g).

C. Safety considerations using hydrogen gas are minimal.

D. It is easily found in nature as an element.

It’s now possible to buy cars that run on hydrogen gas as a fuel; what advantage(s) does hydrogen have as a fuel?

A. It is easily stored as a gas for a long time.

B. Its product of combustion is only H2O(g).

C. Safety considerations using hydrogen gas are minimal.

D. It is easily found in nature as an element.

10/02/2016

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The table at the right lists some specific heat

capacities. Which of the following substances

requires the most heat to increase a 5 gram

sample from 5°C to 12°C?

A) P4

B) Hg

C) Mg

D) Cu

E) Al

The table at the right lists some specific heat

capacities. Which of the following substances

requires the most heat to increase a 5 gram

sample from 5°C to 12°C?

A) P4

B) Hg

C) Mg

D) Cu

E) Al

Would the same apply to molar heat capacity?

10/02/2016

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+

o -1

f

NaCl(s) Na (aq) + Cl (aq)

H / kJ mol 411 240 167

The heats of formation for each of the components

involved in the dissolution of sodium chloride are given

below.H2O

What is the heat of dissolution of NaCl?

A. 4 kJ mol-1

B. -4 kJ mol-1

C. 668 kJ mol-1

D. -668 kJ mol-1

E. none of the above

+

o -1

f

NaCl(s) Na (aq) + Cl (aq)

H / kJ mol 411 240 167

The heats of formation for each of the components

involved in the dissolution of sodium chloride are given

below.H2O

What is the heat of dissolution of NaCl?

A. 4 kJ mol-1

B. -4 kJ mol-1

C. 668 kJ mol-1

D. -668 kJ mol-1

E. none of the above

10/02/2016

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Which of the following statements are correct if the surroundings absorb 145 J of heat from the system while doing 98 J of work on the system?

• I. The system loses 145 J of energy and the surroundings gain 98 J of energy.

• II. The surroundings absorb heat, therefore q is positive.

• III. The system loses 47 J of energy and the surroundings gain 47 J of energy..

• IV. The system gains 47 J of energy and the surroundings lose 47 J of energy.

• V. Work is done on the system, therefore w is positive

A. I & II B. II & III C. II, III, V D. III & V E. IV & V

Which of the following statements are correct if the surroundings absorb 145 J of heat from the system while doing 98 J of work on the system?

• I. The system loses 145 J of energy and the surroundings gain 98 J of energy.

• II. The surroundings absorb heat, therefore q is positive.

• III. The system loses 47 J of energy and the surroundings gain 47 J of energy..

• IV. The system gains 47 J of energy and the surroundings lose 47 J of energy.

• V. Work is done on the system, therefore w is positive

A. I & II B. II & III C. II, III, V D. III & V E. IV & V

10/02/2016

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The standard enthalpy of formation of NiSO4(s) at 25 °C is -872.9 kJ/mole. The chemical equation to which this value applies is

• A. ½ Ni(s) + ½ S(s) + ½ O2(g) → ½ NiSO4(s)

• B. Ni(s) + S(s) + 4 O(g) → NiSO4(s)

• C. Ni(s) + S(s) + 2 O2(g) → NiSO4(s)

• D. Ni(s) + ½ S2(s) + ½ O2(g) → NiSO4 (s)

The standard enthalpy of formation of NiSO4(s) at 25 °C is -872.9 kJ/mole. The chemical equation to which this value applies is

• A. ½ Ni(s) + ½ S(s) + ½ O2(g) → ½ NiSO4(s)

• B. Ni(s) + S(s) + 4 O(g) → NiSO4(s)

• C. Ni(s) + S(s) + 2 O2(g) → NiSO4(s)

• D. Ni(s) + ½ S2(s) + ½ O2(g) → NiSO4 (s)

10/02/2016

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Calculate the enthalpy of formation for gaseous HCl from the following data:

• N2(g) + 4 H2(g) + Cl2(g) → 2 NH4Cl(s) Hf = - 628.8 kJ (1)

• NH3(g) + HCl(g) → NH4Cl(s) Hf = - 176.2 kJ (2)

• The enthalpy of formation of ammonia gas is known to be 46.19 kJ mol-1

Solution:

a) Write down the equation for the formation of ammonia; call this (3)

b) Write down the equation for the formation of HCl (g); call this (4)

c) Add and subtract equations (1)-(3) to get equation (4).

d) Combine the H values to get the answer: -628.8 + (2)(176.2) + (2)(46.19)

= -184.0 kJ for two moles of HCl, so f(HCl) = -92.01 kJ/mol