Question

2

3 of 6 - SCH4U
Date: Name(s):
4. HCl is a corrosive colourless gas that dissolves readily in water.
Aqueous HCl reacts with NaOH to form water and NaCl. In a simple
calorimeter, a 100.00 mL sample of 0.415 mol/L HCl(aq) is mixed
with 50.00 mL of excess NaOH(aq). During the reaction, there is a
rise in temperature by 4.83 °C. Calculate the molar enthalpy change
(in kJ/mol) for the above reaction. SHOW ALL YOUR WORK.

Answer 1

Answer: The molar enthalpy change is 73.04 kJ/mol

Step-by-step explanation:

`HCl+NaOH\rightarrow NaCl+H_2O`

moles of HCl=
`molarity* {\text {vol in L}}=0.415mol/L* 0.1=0.0415mol`

As NaOH is in excess 0.0415 moles of HCl reacts with 0.0415 moles of NaOH.

volume of water = 100.0 ml + 50.0 ml = 150.0 ml

density of water = 1.0 g/ml

mass of water =
`volume * density=150.0ml* 1.0g/ml=150.0g`

`q=m* c* \Delta T`

q = heat released

m = mass = 150.0 g

c = specific heat =
`4.184J/g^0C`

`\Delta T` = change in temperature =
`4.83^0C`

`q=150.0* 4.184* 4.83`

`q=3031.3J`

Thus 0.0415 mol of HCl produces heat = 3031.3 J

1 mol of HCL produces heat =
`(3031.3)/(0.0415)* 1=73043.3J=73.04kJ`

Thus molar enthalpy change is 73.04 kJ/mol