Respuesta :
Given:
Initial temperature of calorimeter T1 = 25 C
Final temperature T2 = 29 C
Heat capacity of calorimeter c = 4.90 kJ/C
Mass of sucrose (m) = 3.5 g
Molar mass of sucrose (M) = 342.3 g/mol
To determine:
The energy change for the combustion of sucrose in kJ/mol
Explanation:
The change in temperature of calorimeter = T2 -T1 = 29 - 25 = 4 C
Heat gained by the calorimeter when the temperature changes by 4 C is-
= 4 C * 4.90 kJ/ 1 C = 19. 6 kJ
Now,
Heat gained by calorimeter = heat lost during combustion of 3.5 g sucrose = -19.6 kJ
# moles of sucrose = 3.5 g/342.3 g.mole-1 = 0.01022 moles
Energy change for the combustion reaction = -19.6/0.01022 = -1917.8 kJ/mol
Ans: Hence the energy change is -1918 kJ/mol
The heat of combustion for the reaction was 1921.5 kJ/mol.
The heat of combustion is the energy for burning of 1 mole of a substance.
[tex]\rm \Delta\;E[/tex] = c[tex]\Delta[/tex]T
C = specific heat of calorimeter = [tex]\rm 4.90\;kJ/g^\circ C[/tex]
[tex]\Delta[/tex]T = change in temperature
[tex]\rm \Delta T\;=\;T_f_i_n_a_l\;-\;T_i_n_i_t_i_a_l[/tex]
[tex]\Delta[/tex]T = 29 - [tex]\rm 25^\circ C[/tex]
[tex]\rm \Delta T\;=\;4\;^\circ C[/tex]
[tex]\Delta[/tex] E = 4.90 [tex]\times[/tex] 4 kJ
[tex]\rm \Delta E[/tex] = 19.6 kJ
Molar heat of combustion of sucrose = [tex]\rm \frac{\Delta E}{moles}[/tex]
moles of sucrose = [tex]\rm \frac{weight}{molecular\;weight}[/tex]
moles of sucrose = [tex]\rm \frac{3.50}{342.3}[/tex] moles
moles of sucrose = 0.0102 moles
[tex]\rm \Delta E_e_x_n[/tex] = [tex]\rm \frac{19.6}{0.0102}[/tex] kJ
[tex]\rm \Delta E_e_x_n[/tex] = 1921.5 kJ/mol
[tex]\rm \Delta E_e_x_n[/tex] = 1921.5 kJ/mol.
For more information about the heat of combustion, refer the link:
https://brainly.com/question/14317568?referrer=searchResults
