Question

In a coffee-cup calorimeter, 150.0 mL of 0.50 M HCl is added to 50.0 mL of 1.00 M NaOH to make 200.0 g solution at an initial temperature of 48.2°C. If the enthalpy of neutralization for the reaction between a strong acid and a strong base is −56 kJ/mol, calculate the final temperature of the calorimeter contents. Assume the specific heat capacity of the solution is 4.184 J°C⁻¹ g⁻¹ and assume no heat loss to the surroundings.

Answer 1

Answer:

51.54°C the final temperature of the calorimeter contents.

Explanation:

$HCl+NaOH\rightarrow H_2O+NaCl,\Delta H=-56 kJ/mol$

$moles=Molarity\times Volume (L)$

Molarity of HCl= 0.50 M

Volume of HCl= 150.0 mL = 0.150 L

Moles of HCl= n

$n=0.50 M\times 0.150 L=0.075 mol$

Molarity of NaOH= 1.00 M

Volume of NaOH= 50.0 mL = 0.050 L

Moles of NaOH= n'

$n'=1.00 M\times 0.050 L=0.050 mol$

Since moles of NaOH are less than than moles of HCl. so energy release will be for neutralization of 0.050 moles of naOH by 0.050 moles of HCl.

n = 0.050

$-56 kJ/mol=-\frac{Q}{n}$

$Q=56 kJ/mol\times 0.050 kJ/mol=2.8 kJ=2800 J$

(1 kJ= 1000 J)

The energy change released during the reaction = 2800 J

Volume of solution = 150.0 mL + 50.0 mL = 200.0 mL

Density of the solution (water) = 1.00g/mL

Mass of the solution , m= 200 mL × 1.00 g/mL = 200 g

Now , calculate the final temperature by the solution from :

$q=mc\times (T_{final}-T_{initial})$

where,

q = heat gained = 2800 J

c = specific heat of solution = $4.184 J/^oC$

$T_{final}$ = final temperature = $?$

$T_{initial}$ = initial temperature = $48.2^oC$

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

$2800 J=200.0 g\times 4.184 J/^oC\times (T_{final}-48.2)^oC$

$T_{final}= 51.54^oC$

51.54°C the final temperature of the calorimeter contents.