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

51.54°C the final temperature of the calorimeter contents.

Step-by-step explanation:

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

`moles=Molarity* 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* 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* 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=-(Q)/(n)`

`Q=56 kJ/mol* 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* (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* 4.184 J/^oC* (T_(final)-48.2)^oC`

`T_(final)= 51.54^oC`

51.54°C the final temperature of the calorimeter contents.