Chemistry: Unit F322: Chains, Energy and Resources



Chemistry: Unit F322: Chains, Energy and Resources

Moles & Enthalpy

Molar mass is the mass per mole of a substance. A mole of any substance contains 6.03 x 1023 atoms. For elements, the molar mass is simply the relative atomic mass in g mol-1. To obtain the molar mass of a molecule or compound, you must add together the relative atomic masses of each atom in the formula.

The number of moles of a substance can be calculated using the formula:

Number of Moles, n = Mass

Molar Mass

The concentration of a solution is the amount of solute, in moles, dissolved in each dm3 (1000cm3) of solution. The number of moles in a solution can be calculated using this formula:

Number of moles = Concentration x Volume in dm3

In order to obtain the volume in dm3 from cm3, divide the volume in cm3 by 1000, so the formula becomes:

Number of moles = Concentration x Volume in cm3

1000

Enthalpy Changes: Enthalpy is the heat energy that is stored in a chemical system. An enthalpy change, ΔH, is the heat energy exchange with the environment at constant pressure.

Determination of Enthalpy Change using Calorimetry

The heat energy transferred in a reaction, q (IN JOULES), can be found using this equation:

Q = m c Δt

where m = mass of water, c = specific heat capacity and Δt is the temperature change.

The enthalpy change of a reaction can be found by:

ΔH = q

n

Step 1: Calculate the energy transfer using q = mcΔt:

m=75 (1cm3 of water weighs 1g)

c = 4.2

t= 12.5

Step 2: Calculate the enthalpy change using ΔH = q/n

q = -3.9375 kJ (negative because the reaction was exothermic, and we have divided b 1000 to get q in kilojoules rather than joules)

n = number of moles of reactant:

number of moles = mass/ molar mass

= 0.15/46 (the molar mass of ethanol)

n= 0.0036

Step 1: Calculate the energy transfer using q = mcΔt:

m=35 (1cm3 of water weighs 1g)

c = 4.2

t= 8.5

Step 2: Calculate which reactant is not in excess:

No mol NaHCO3 = Mass/Mr = 3.71/84 = 0.044 mol

No mol HCl = Conc x Vol/1000 = 1.5 x 0.035 = 0.0525

The NaHCO3 is not in excess, and this therefore is the value of n which will be used in ΔH = q/n.

Step 3: Calculate the enthalpy change using ΔH = q/n

q = 1249.5 / 1000 = 1.2495 kJ

n = 0.044 moles

Standard enthalpy changes enable scientists to compare the enthalpy changes of different reactions. A standard enthalpy change must be found using standard conditions.

ΔH Ѳ refers to an enthalpy change under standard conditions. These conditions are;

▪ A pressure of 100 kPa

▪ A stated temperature, usually 298 K (25˚C) used

▪ A concentration of 1 moldm-3 for aqueous solutions

Standard Enthalpy change of Reaction: ΔH( r

The enthalpy change when the amounts of reactants as shown in the reaction equation react together under standard conditions, all reactants and products being in their standard states.

Standard Enthalpy change of Combustion: ΔH( c

The enthalpy change that takes place when one mole of substance reacts completely with oxygen under standard conditions, all reactants and products being in their standard states.

Standard Enthalpy change of Formation: ΔH( f

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

Hess’ Law states that the enthalpy change of a reaction depends only on its initial and final states, and is independent of the route taken. The enthalpy change of the indirect route = the enthalpy change of the direct route.

Calculating Enthalpy Change from Combustion data

1. Write the balanced symbol equation of what you want to find across the top

2. Write the oxide combustion products at the bottom (H2O and CO2)

3. Work out reaction route – ARROWS POINT DOWNWARDS

|Substance |ΔHѲc |

|Carbon, C |-394 kJmol-1 |

|Hydrogen, H2 |-286 kJmol-1 |

|Methane, CH4 |-890 kJmol-1 |

Calculating Enthalpy Change from Formation data

The enthalpy of a reaction = the enthalpy change of the formation of the products - the enthalpy change of the formation of the reactants.

|Substance |ΔHѲf |

|Methane, CH4 |-75 kJmol-1 |

|Water, H2O |-242 kJmol-1 |

|Carbon Monoxide, CO |-110 kJmol-1 |

ΔHr = ΣΔHf (products) - ΣΔHf (reactants)

ΔHr = (-110) – [-75+(-242)]

ΔHr = 207 kJmol-1

Average Bond Enthalpy is the average enthalpy change when breaking 1 mole of bonds in a compound which is in its gaseous state.

|Bond |Average Bond Enthalpy |

|C-H |413 |

|O=O |497 |

|C=O |740 |

|O-H |463 |

Breaking Bonds is an endothermic process. It has a positive sign, as energy must be put into the system to break the chemical bonds.

Making bonds in an exothermic process. It has a negative sign as energy is released from the system.

Calculating Enthalpy Change from average bond enthalpies

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The molar mass of carbon is 12 g mol-1.

The molar mass of oxygen is 16 g mol-1.

The molar mass of CO2 is 12 + (16 x 2) = 44 g mol-1

Reactants

Products

Enthalpy, H

Reaction Pathway

ΔH +ve

ΔH -ve

Reaction Pathway

Enthalpy, H

Products

Reactants

In an exothermic reaction, ΔH is negative, because heat is given out to the surroundings so the reacting chemicals lose energy.

In an endothermic reaction, ΔH is positive, because heat is taken in from the surroundings so the reacting chemicals gain energy.

The first law of thermodynamics states that energy may be exchanged between a chemical system and the surroundings but the total energy remains constant. Energy cannot be created nor destroyed, only changed from one form into another.

Q1) The combustion of 0.15g of ethanol (C2H5OH) in a spirit burner increased the temperature of 75cm3 of water by 12.5˚c. Calculate the enthalpy of combustion for ethanol.

q = 75 x 4.2 x 12.5

q = 3937.5 joules

where q= heat energy transferred IN KILOJOULES and n= the number of moles of reactant not in excess.

ΔH = q/n

= -3.9375 / 0.0036

= -1093.75 kJmol-1

Q2) A student added 3.71g of NaHCO3 to 35cm3 of 1.5 moldm-3 HCl(aq). The temperature fell by 8.5˚. Calculate the enthalpy change for the reaction.

q = 35 x 4.2 x 8.5

q = 1249.5 joules

ΔH = q/n

= 1.2495 / 0.044

= 28.4 kJmol-1

H2 + ½ O2 H2 O [ΔH(r =-286 kJmol-1]

CH4 + 2O2 CO2 + 2H2 O [ΔH( c =-890 kJmol-1]

2Na + C + 1 ½ O2 Na2CO3 [ΔH( f = -1131 kJmol-1]

A

B

C

ΔH1

ΔH2

ΔH3

Direct Route = Indirect Route

ΔH1 = ΔH2 + ΔH3

Calculate the standard enthalpy change of the formation of methane, using the values in the table.

C + 2H2 CH4

CO2 + 2H2O

ΔH1

ΔH2

2ΔH3

ΔH4

ΔH1 = ΔH2 + 2ΔH3 – ΔH4

ΔH1 = [-394 + (2 x -286)] – (-890)

ΔH1 = -76 kJmol-1

ΔHr = ΣΔHf (products) - ΣΔHf (reactants)

Calculate the enthalpy change for the reaction between methane and steam, producing carbon monoxide and hydrogen.

CH4 + H2O CO + 3H2

ΔHr = Σ Bonds Broken + Σ Bonds formed

Calculate the standard enthalpy change of the combustion of methane using average bond enthalpies.

CH4 + 2O2 CO2 + 2H2O

Endothermic – Bonds are broken, so energy is taken in (+ve)

4 x C-H = 413x4

2 x O=O = 2x497

= +2646

Exothermic – Bonds are formed, so energy is given out (-ve)

2 x C=O = 2 x 740

4 x O-H = 4 x 463

= - 3332

2646 + (-3332)

= -686 kJmol-1

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