EDEXCEL IGCSE CHEMISTRY: DOUBLE SCIENCE 复习笔记:3.1.3 Energetics Calculations

EDEXCEL IGCSE CHEMISTRY: DOUBLE SCIENCE 复习笔记:3.1.3 Energetics Calculations

Calculate Heat Energy Change

 

  • In order the calculate heat energy changes you need to know the mass of the substance being heated, the temperature change and the specific heat capacity of the substance
  • The  specific heat capacity (c) is the energy needed to raise the temperature of 1 g of a substance by 1 K
  • The specific heat capacity of water is 4.18 J g-1 K-1
  • The energy transferred as heat can be calculated by:

 

5.1.3-Calorimetry-variables-symbols_1 

Equation for calculating energy transferred in a calorimeter

 

  • The temperature change in Kelvin is the same as the temperature change in degrees Celsius

 

 

Worked Example

Excess iron powder was added to 100.0 cmof 0.200 mol dm-3  copper(II)sulfate solution in a calorimeter. The reaction equation was as followsFe (s) + CuSO(aq)     FeSO(aq) + Cu (s)The maximum temperature rise was 7.5 oC. Determine the heat energy change of the reaction, in kJ

 

Answer:

 

The solution is assumed to have the same density as water, so 100.0 cm has a mass of 100 g

 

q = m x c x ΔT

 

q = 100 g x 4.18 J g-1 K-1 x 7.5 K = – 3135 J = -3.13 kJ

 

 

Worked Example

1.023 g of propan-1-ol (M = 60.11 g mol-1) was burned in a spirit burner and used to heat 200 g of water in a copper calorimeter. The temperature of the water rose by 30 oC.Calculate the heat energy change for the combustion of propan-1-ol using this data.

 

Answer:

 

Calculate q

 

q = m x c x ΔT

 

q = 200 g x 4.18 J g-1 K-1 x 30 K = – 25 080 J = -25 kJ

 

 

Calculate Molar Enthalpy Change

 

  • Molar enthalpy change is the heat energy change per mole of substance
  • The symbol is ΔH and it has the unit kJ per mole
  • If is found by first determining the heat energy change for the reaction, q, and then dividing by the number of moles, n, of the substance

 

molar enthalpy change  = heat change for the reaction ÷ number of moles

 

ΔH = q ÷ n 

 

Worked Example

The energy from 0.01 mol of propan-1-ol was used to heat up 250 g of water. The temperature of the water rose from 298 K to 310 K (the specific heat capacity of water is 4.18 J g-1 K-1.Calculate the molar enthalpy of combustion.

 

Answer:

 

Step 1: q = m x c x ΔT

 

(of water) = 250 g

 

(of water) = 4.18 J g-1 K-1

 

Δ(of water) = 310 – 298 K

 

= 12 K

 

Step 2: q = 250 x 4.18 x 12

 

= 12 540 J

 

Step 3:  This is the energy released by 0.01 mol of propan-1-ol

 

Total energy    ΔH = q ÷ n = 12 540 J ÷ 0.01 mol = 1 254 000 J mol-1

 

Total energy = – 1254 kJ mol-1

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