Respuesta :
Answer: [tex]4403kJ/mole[/tex]
Explanation:
First we have to calculate the heat absorbed by bomb calorimeter
Formula used :
[tex]q_b=c_b\times (T_{final}-T_{initial})[/tex]
[tex]q_b[/tex] = heat absorbed by calorimeter = ?
[tex]c_b[/tex] = specific heat of = 10.69 kJ/K
[tex]T_{final}[/tex] = final temperature = [tex]44.688^oC=(273+44.688)K=317.688K[/tex]
[tex]T_{initial}[/tex] = initial temperature = [tex]20.486^oC=(273+20.486)K=293.486K[/tex]
[tex]q_b=10.69kJ/K\times (317.688-293.486)=258.7kJ[/tex]
As heat absorbed by calorimeter is equal to the heat released by acetylene during combustion.
Thus 1.53 gram of acetylene releases heat of combustion = 258.7kJ
So, 26.04 g/mole of acetylene releases heat of combustion [tex]\frac{258.7}{1.53}\times 26.04=4403kJ/mole[/tex]
Therefore, the heat of combustion of acetylene is, [tex]4403kJ/mole[/tex]
The ΔE (in kJ/mol) for this combustion reaction is 4385 kJ/mol
We know that the number of moles = mass/molar mass
where;
- molar mass of acetylene = 26.04 g/mol
From the given information, the number of moles of acetylene used:
= 1.53 g/26.04 g/mol
= 0.059 mol
When a given mass of fuel is burned in a bomb calorimeter with a known heat capacity, we can estimate the heat evolved during combustion by measuring the temperature change.
∴
Heat evolved = heat capacity × change in temperature
Heat evolved = 10.69 kJ/K × (44.688 - 20.486)K
Heat evolved = 10.69 kJ/K × 24.202K
Heat evolved = 258.719 kJ
We know that the heat of combustion ΔE is the amount of heat evolved by the complete combustion of 1 mole of a gas.
∴
[tex]\mathbf{\Delta E = \dfrac{heat \ evolved}{n}}[/tex]
[tex]\mathbf{\Delta E = \dfrac{258.719 \ kJ}{0.059 mol}}[/tex]
[tex]\mathbf{\Delta E = 4385 \ kJ/mol}[/tex]
Learn more about the heat of combustion here:
https://brainly.com/question/10093007?referrer=searchResults
