measuring changes in chemical reactions - U3 AOS 1

Question 1

mass to mass stoichiometry

Answer 1

• determine molar mass of known and unknown
• convert known substance mass to mole
• use stoichiometry rations to find mole of unknown
• n(unknown) = unknown/known x n(known)

Question 2

mass to volume stoichiometry

Answer 2

• molar volume of gases is 24.8
• measured in L mol-1
• n=v/Vm
• density = mass/volume

Question 3

volume to volume stoichiometry

Answer 3

• all gases occupy the same volume at equal pressures and temperature
• v(known) = unknown/known x v(known)

Question 4

enthalpy changes

Answer 4

• energy released in kj mol-1 = molar heat of combustion x n(fuel)
• energy released in kj g-1 = heat o combustion x m(fuel)
• ensure to work out energy mole when one mole of fuel is combusted

Question 5

combustion and greenhouse gases

Answer 5

• combustion of carbon based fuels produces carbon dioxide, a greenhouse has which can absorb infrared radiation
• carbon dioxide contributes to increased global warming an the greenhouse effect
• methane also contributes to the global warming effect

Question 6

reducing the effect of greenhouse gases

Answer 6

• capturing methane gas from agriculture does increase the net volume of CO2 in the atmosphere but it reduces the effect of methane which is worse

Question 7

specific heat capacity

Answer 7

• the energy needed in joules to increase the temperature of 1 gram of a substance by 1 degrees
• water has a high specific heat capacity due to hydrogen bonding between water molecules
• high specific heat capacity means the substance is good a storing heat.
• measured in joules per gram per degrees

Question 8

specific heat capacity formular

Answer 8

q (J) = m x c x ΔT

c = 4.18

Question 9

estimating heat released

Answer 9

• a vessel can be filled with water and the mass and initial temp recorded. Once the fuel is combusted, the specific heat capacity of water can be calculated.
• heat of combustion = q/m(fuel)
• molar heat of combustion (kj/mol) = q/n(fuel)
• chance from j to kj

Question 10

calorimetry

Answer 10

• the use of a calorimeter to estimate the enthalpy change of a reaction that occurs in solutions

Question 11

calorimeter

Answer 11

• an insulated vessel designed to minimise and account for heat loss

Question 12

calibration factor

Answer 12

• the energy needed to be released/absorbed in a given calorimeter to increase/decrease water by 1 degrees
• Calibration allows greater accuracy than using the heat capacity of water alone, as it
  takes into account heat absorbed by the entire inside of the calorimeter.
• joules per degrees

Question 13

chemical calibration method

Answer 13

• conduct an exothermic reaction with a known enthalpy change and amount of fuel, them measure the ΔTof the water. .
• q (kj) = n(fuel) x heat of combustion
• calibration factor = q/ΔT

Question 14

electrical calibration method

Answer 14

• Electrical calibration is achieved by calibrating the calorimeter using an electrically heated coil to supply a measured quantity of electrical energy, which is converted to heat energy. The heat energy
  is transferred to a known mass of a substance, usually water, and then the temperature rise is measured.
• energy released (q) in joules = v x l x t
• v = potential difference
• I = current
• t = seconds
• CF = q/ΔT
• CF = vlt/ΔT

Question 15

using a calibration factor

Answer 15

1. conduct the unknown reaction in the same calorimeter
2. measure the ΔT of the water
3. calculate the energy using q(J) = CF x ΔT
4. q can be used to calculate the enthalpy change of the reaction.

Question 16

heat loss

Answer 16

• when heat form the burning fuel is lost into the surrounding air, the temperature does not increase as much as it would have, resulting in a lower ΔT

Question 17

energy content of food

Answer 17

energy content of food = energy transfer/ change in mass of fuel

Question 18

energy transformations

Answer 18

• energy is converted from one form to another

Question 19

energy transformation efficiency

Answer 19

energy efficiency percentage = experimental value/theoretical value x 100

energy loss = 100%-energy efficiency%

Question 20

energy

Answer 20

energy available = mass in sample x heat of combustion

Question 21

why are lipids metabolised first

Answer 21

they have less oxygen in their structure and so more C-H bonds which are energy rich meaning they will have a higher energy content per gram

lipids are more reduced whereas carbohydrates are already partially oxidised. thus means carbohydrates yield less energy per gram