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

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

Step-by-step 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* V=0.250 M* 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* V=0.250 M* 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=-(q)/(n)`

`q=\Delta H* n=-57.2 kJ/mol * 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* volume=d* v=1 g/ml * 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* (T_(f)-T_(i))`

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

`715 J=100 g* 4.18J/^oC* (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.