Respuesta :
Answer:
22.48°C is the final temperature of the solution.
Explanation:
Heat of neutralization of reaction , when 1 mol of nitric acid reacts= ΔH= -56.2 kJ/mol= -56200 J/mol
[tex]Moles (n)=Molarity(M)\times Volume (L)[/tex]
Moles of nitric acid = n
Volume of nitric acid solution = [tex]8.10\times 10^2 mL= 0.81L[/tex]
Molarity of the nitric acid = 0.600 M
[tex]n=0.600 M\times 0.81 L=0.486 mol[/tex]
Moles of barium hydroxide = n'
Volume of barium hydroxide solution = [tex]8.10\times 10^2 mL= 0.81L[/tex]
Molarity of the barium hydroxide= 0.300 M
[tex]n'=0.300 M\times 0.81 L=0.243 mol[/tex]
[tex]2HNO_3+Ba(OH)_2\rightarrow Ba(NO_3)_3+2H_2O[/tex]
According to reaction, 2 mol of nitric acid reacts with 1 mol of barium hydroxide .Then 0.486 mol of nitric acid will react with :
[tex]\frac{1}{2}\times 0.486 mol=0.243 mol[/tex] barium hydroxide.
Heat release when 0.486 mol of nitric acid reacted = Q
= ΔH × 0.486 = -56200 J/mol × 0.486 mol=-27,313.2 J
Heat absorbed by the mixture after reaction = Q' = -Q = 27,313.2 J
Volume of the nitric solution = 0.81 L = 810 mL
Volume of the ferric nitrate solution = 0.81 L = 810 mL
Total volume of the solution = 810 mL + 810 mL = 1620 mL
Mass of the final solution = m
Density of water = density of the final solution = d = 1 g/mL
[tex]Mass=density\times Volume[/tex]
[tex]m=1 g/ml\times 1620 ml=1620 g[/tex]
Initial temperature of the both solution were same = [tex]T_1=18.46^oC[/tex]
Final temperature of the both solution will also be same after mixing= [tex]T_2[/tex]
Heat capacity of the mixture = c = 4.184 J/g°C
Change in temperature of the mixture = ΔT =[tex] (T_2-T_1)[/tex]
[tex]Q=mc\Delta T=mc(T_2-T_1)[/tex]
[tex] 27,313.2 J= 1620 g\times 4.184 J/g^oC\times (T_2-18.46^oC)[/tex]
[tex]T_2=22.49^oC[/tex]
22.48°C is the final temperature of the solution.
