Question

What will be the final temperature of the solution in a coffee cup calorimeter if a 50.00 mL sample of 0.250 M HCl(aq) is added to a 50.00 mL sample of 0.250 NaOH(aq). The initial temperature is 19.50 °C and the Hrxn is −57.2 kJ/mol NaOH. (assume the density of the solution is 1.00 g/mL and the specific heat of the solution is 4.18 J/g∙°C)

Answer 1

Answer:

21.21°C will be the final temperature of the solution in a coffee cup calorimeter.

Explanation:

$HCl+NaOH\rightarrow H_2O+NaCl$

$\Delta H$ = enthalpy change = -57.2 kJ/mol of NaOH

Moles of sodium hydroxide = n

Molarity of the NaOH = 0.250 M

Volume of NaOH solution = V = 50.00 mL = 0.050 L

$n=Molarity\times V=0.250 M\times 0.050 L= 0.0125 mol$

Moles of HCl = n'

Molarity of the HCl= 0.250 M

Volume of HCl solution = V' = 50.00 mL = 0.050 L

$n'=Molarity\times V=0.250 M\times 0.050 L= 0.0125 mol$

Since 1 mole of Hcl reacts with 1 mole of NaoH. Then 0.0125 mole of HCl will react with 0.0125 mole of NaOH.

The enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

$q=\Delta H\times n=-57.2 kJ/mol \times 0.0125 mol= -0.715 kJ=-715 J$

q = heat released on reaction= -715 J

now, we calculate the heat gained by the solution.:

Q= -q = -(-715 J) = 715 J

m = mass of the solution = ?

Volume of the solution formed by mixing, v = 50.00 ml + 50.00 mL = 100.00 mL

Density of the solution = density of water = d = 1 g/mL

$mass=density\times volume=d\times v=1 g/ml \times 100.00 ml=100 g$

m = 100 g

q = heat gained = ?

c = specific heat = $4.18 J/^oC$

$T_{f}$ = final temperature = ?

$T_{i}$ = initial temperature = $19.50^oC$

$Q=mc\times (T_{f}-T_{i})$

Now put all the given values in the above formula, we get:

$715 J=100 g\times 4.18J/^oC\times (T_f-19.50)^oC$

$T_f=21.21 ^oC$

21.21°C will be the final temperature of the solution in a coffee cup calorimeter.