Question

A quantity of 8.10 × 102 mL of 0.600 M HNO3 is mixed with 8.10 × 102 mL of 0.300 M Ba(OH)2 in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of both solutions is the same at 18.46°C. The heat of neutralization when 1.00 mol of HNO3 reacts with 0.500 mol Ba(OH)2 is −56.2 kJ/mol. Assume that the densities and specific heats of the solution are the same as for water (1.00 g/mL and 4.184 J/g · °C, respectively). What is the final temperature of the solution?

Answer 1

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

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of nitric acid = n

Volume of nitric acid solution = $8.10\times 10^2 mL= 0.81L$

Molarity of the nitric acid = 0.600 M

$n=0.600 M\times 0.81 L=0.486 mol$

Moles of barium hydroxide = n'

Volume of barium hydroxide solution = $8.10\times 10^2 mL= 0.81L$

Molarity of the barium hydroxide= 0.300 M

$n'=0.300 M\times 0.81 L=0.243 mol$

$2HNO_3+Ba(OH)_2\rightarrow Ba(NO_3)_3+2H_2O$

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 :

$\frac{1}{2}\times 0.486 mol=0.243 mol$ 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

$Mass=density\times Volume$

$m=1 g/ml\times 1620 ml=1620 g$

Initial temperature of the both solution were same = $T_1=18.46^oC$

Final temperature of the both solution will also be same after mixing= $T_2$

Heat capacity of the mixture = c = 4.184 J/g°C

Change in temperature of the mixture = ΔT =$(T_2-T_1)$

$Q=mc\Delta T=mc(T_2-T_1)$

$27,313.2 J= 1620 g\times 4.184 J/g^oC\times (T_2-18.46^oC)$

$T_2=22.49^oC$

22.48°C is the final temperature of the solution.