Respuesta :
Answer :
(a) The heat capacity of calorimeter [tex]31.25kJ/^oC[/tex]
(b) The energy of combustion of acetaldehyde is, 1109.8 kJ/mol
Explanation :
First we have to calculate the heat produced.
[tex]\Delta H=\frac{q}{n}[/tex]
where,
[tex]\Delta H[/tex] = enthalpy change = -803 kJ/mol
q = heat released = ?
m = mass of [tex]CH_4[/tex] = 6.91 g
Molar mass of [tex]CH_4[/tex] = 16 g/mol
[tex]\text{Moles of }CH_4=\frac{\text{Mass of }CH_4}{\text{Molar mass of }CH_4}=\frac{6.91g}{16g/mole}=0.432mole[/tex]
Now put all the given values in the above formula, we get:
[tex]-803kJ/mol=\frac{q}{0.432mole}[/tex]
[tex]q=-346.896kJ[/tex]
(a) Now we have to calculate the heat capacity of calorimeter.
[tex]q=c\times (\Delta T)[/tex]
where,
q = heat produced = 346.896 kJ = 346896 J
c = heat capacity of calorimeter = ?
[tex]\Delta T[/tex] = change in temperature = [tex]11.1^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]346896J=c\times (11.1^oC)[/tex]
[tex]c=31251.8J^oC=31.25kJ/^oC[/tex]
(b) Now we have to calculate the moles of acetaldehyde.
Mass of [tex]CH_3CHO[/tex] = 14.0 g
Molar mass of [tex]CH_3CHO[/tex] = 44 g/mol
[tex]\text{Moles of }CH_3CHO=\frac{\text{Mass of }CH_3CHO}{\text{Molar mass of }CH_3CHO}=\frac{14.0g}{44g/mole}=0.3182mole[/tex]
Now we have to calculate the heat produced in combustion.
[tex]q=c\times (\Delta T)[/tex]
where,
q = heat produced = ?
c = heat capacity of calorimeter = [tex]31.25kJ/^oC[/tex]
[tex]\Delta T[/tex] = change in temperature = [tex]11.3^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]q=31.25kJ/^oC\times (11.3^oC)[/tex]
[tex]q=353.125kJ[/tex]
Now we have to calculate the energy of combustion of acetaldehyde.
[tex]\Delta H=\frac{q}{n}[/tex]
where,
[tex]\Delta H[/tex] = enthalpy change = ?
q = heat released = 353.125 kJ
n = moles of [tex]CH_3CHO[/tex] = 0.3182mole
Now put all the given values in the above formula, we get:
[tex]\Delta H=\frac{353.125 kJ}{0.3182mole}[/tex]
[tex]\Delta H=1109.8kJ/mol[/tex]
