Question

In a constant‑pressure calorimeter, 70.0 mL of 0.310 M Ba ( OH ) 2 was added to 70.0 mL of 0.620 M HCl . The reaction caused the temperature of the solution to rise from 21.12 ∘ C to 25.34 ∘ C. If the solution has the same density and specific heat as water, what is heat absorbed by the solution? Assume that the total volume is the sum of the individual volumes. (And notice that the answer is in kJ).

Answer 1

Answer: The amount of heat absorbed by the solution is 56.98 kJ

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$    .....(1)

• For Barium hydroxide:

Molarity of barium hydroxide solution = 0.310 M

Volume of solution = 70 mL = 0.070 L   (Conversion factor:  1 L = 1000 mL)

Putting values in equation 1, we get:

$0.310M=\frac{\text{Moles of }Ba(OH)_2}{0.070L}\\\\\text{Moles of }Ba(OH)_2=(0.310mol/L\times 0.070L)=0.0217mol$

• For HCl:

Molarity of HCl solution = 0.620 M

Volume of solution = 70 mL = 0.070 L

Putting values in equation 1, we get:

$0.620M=\frac{\text{Moles of HCl}}{0.070L}\\\\\text{Moles of HCl}=(0.620mol/L\times 0.070L)=0.0434mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O$

By Stoichiometry of the reaction:

2 moles of HCl produces 2 moles of water

So, 0.0434 moles of HCl will produce = $\frac{2}{2}\times 0.0434=0.0434moles$ of water

• To calculate the mass of solution, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1 g/mL

Volume of solution = [70 + 70] = 140 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of solution}}{140mL}\\\\\text{Mass of solution}=(1g/mL\times 140mL)=140g$

• To calculate the amount of heat absorbed, we use the equation:

$q=mc\Delta T$

where,

q = heat absorbed

m = mass of solution = 140 g

c = heat capacity of solution= 4.186 J/g°C

$\Delta T$ = change in temperature = $T_2-T_1=(25.34-21.12)^oC=4.22^oC$

Putting values in above equation, we get:

$q=140g\times 4.186J/g^oC\times 4.22^oC=2473.08J=2.473kJ$

To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat absorbed = 2.473 kJ

n = number of moles of water = 0.0434 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction

Putting values in above equation, we get:

$\Delta H_{rxn}=\frac{2.473kJ}{0.0434mol}=56.98kJ/mol$

Hence, the amount of heat absorbed by the solution is 56.98 kJ