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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.

Respuesta :

Answer: The molar enthalpy change is 73.04 kJ/mol

Explanation:

[tex]HCl+NaOH\rightarrow NaCl+H_2O[/tex]

moles of HCl= [tex]molarity\times {\text {vol in L}}=0.415mol/L\times 0.1=0.0415mol[/tex]

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 = [tex]volume \times density=150.0ml\times 1.0g/ml=150.0g[/tex]

[tex]q=m\times c\times \Delta T[/tex]

q = heat released

m = mass  = 150.0 g

c = specific heat = [tex]4.184J/g^0C[/tex]

[tex]\Delta T[/tex] = change in temperature = [tex]4.83^0C[/tex]

[tex]q=150.0\times 4.184\times 4.83[/tex]

[tex]q=3031.3J[/tex]

Thus 0.0415 mol of HCl produces heat = 3031.3 J

1 mol of HCL produces heat = [tex]\frac{3031.3}{0.0415}\times 1=73043.3J=73.04kJ[/tex]

Thus molar enthalpy change is 73.04 kJ/mol

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