Respuesta :
1,41 x 10³ kJ/mol
Further explanation
Given:
- A bomb calorimeter has a heat capacity of C = 2.47 kJ/K.
- A 0.120-g sample of ethylene (C₂H₄) was burned in this calorimeter.
- The temperature increased by ΔT = +2.44 K.
Question:
Calculate the enthalpy change per mole of ethylene combusted.
The Process:
Step-1: find out the amount of heat energy (Q) transferred
We will solve the problem with calorimetry calculations. We all see that the bomb calorimeter has a thermal (or thermal) capacity. During the sample was burned, there was an increase in temperature.
Q = heat capacity x temperature difference
[tex]\boxed{ \ Q = C \ \Delta T \ }[/tex]
[tex]\boxed{ \ Q = 2.47 \ \frac{kJ}{K} \times 2.44 \ K \ }[/tex]
Hence the amount of heat energy transferred in the reaction of the bomb calorimeter is 6.0268 kJ.
Step-2: calculate the number of moles of ethylene combusted
Molar mass of ethylene (C₂H₄) = (2 x 12.011) + (4 x 1.008) = 28.054.
Let us convert the number of grams of a sample of ethylene into moles.
[tex]\boxed{ \ moles (n) = \frac{mass}{molar \ mass} \ }[/tex]
[tex]\boxed{ \ moles (n) = \frac{0.120}{28.054} \ }[/tex]
We get [tex]\boxed{ \ 4.277 \times 10^{-3} \ moles \ }[/tex] of ethylene.
Step-3: calculate the enthalpy change per mole of ethylene combusted
[tex]\boxed{ \ \Delta H = - \frac{heat \ transferred}{moles \ of \ substance}\ \ } \rightarrow \boxed{ \ \Delta H = - \frac{Q}{n} \ }[/tex]
[tex]\boxed{ \ \Delta H = - \frac{6.0268 \ kJ}{4.277 \times 10^{-3} \ moles} \ }[/tex]
Thus the enthalpy change per mole of ethylene combusted is [tex]\boxed{ \ \Delta H = 1409.12 \ kJ/mol \ } \rightarrow \boxed{ \ \Delta H \approx 1,41 \times 10^3 \ kJ/mol \ }[/tex]
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Notes:
- The enthalpy change for a reaction can be measured experimentally by using a calorimeter.
- In a simple calorimeter all the heat evolved in exothermic reaction when is used to raise the temperature of a known mass of water (or heat capacity of a bomb calorimeter).
- For endothermic reactions the heat transferred from the water to the reaction can be calculated by measuring the lowering of temperature of a known mass of water (or heat capacity of a bomb calorimeter).
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The enthalpy change per mole of ethylene combusted is 1401.63 kJ/mol.
The given parameters;
- heat capacity of the bomb, C = 2.47 kJ/k
- increase in temperature, Δt = 2.44 k
- mass of the ethylene sample = 0.12 g
The quantity of heat transferred to the calorimeter when the temperature was increased;
Q = CΔt
Q = (2.47)(2.44)
Q = 6.027 kJ
The number of moles of the ethylene is calculated as follows;
mass of ethylene = (12 x 2) + (1 x 4) = 28 g
[tex]no \ of \ mole = \frac{0.12}{28} = 0.0043 \ mol[/tex]
The enthalpy change per mole of ethylene combusted is calculated as follows;
[tex]\Delta H = \frac{6.027 \ kJ}{0.0043 \ mol} = 1401.62 \ kJ/mol[/tex]
Thus, the enthalpy change per mole of ethylene combusted is 1401.63 kJ/mol.
Learn more here:https://brainly.com/question/13614547