50 mL of 2.2 M HCl is combined with 50 mL of 2.0 M NaOH in a coffee-cup calorimeter. The temperature of the solution increases 13.4°C. Assume the
calorimeter is a perfect insulator and the mixture has a specific heat capacity identical
to pure water (4.184 wc)
.
Calculate the heat change from the chemical reaction (in units of kJ). (Pay attention to
sign conventions and do not write the units in your typed answer.)

-56.1kJ/mol

Explanation:

The reaction between HCl and NaOH is:

NaOH + HCl → NaCl + H₂O + ΔH

Where ΔH is heat change in  the reaction.

As the temperature of the solution increases, the heat is released and ΔH < 0

The heat released in the reaction is obtained using coffe-cup calorimeter equation:

Q = C×m×ΔT

Where Q is heat

C is specific heat of the solution (4.184J/g°C)

m is mass of solution: Assuming density = 1g/mL, 100mL of solution = 100g

And ΔT is change in temperature (13.4°C)

Replacing:

Q = C×m×ΔT

Q = -4.184J/g°C×100g×13.4

Q = -5606.6J

Now, in the reaction you have:

Moles HCl:

0.050L * (2.2mol/L) = 0.11 moles

Moles NaOH:

0.050L * (2.0mol/L) = 0.1 moles

That means the moles of reaction are 0.1 moles, and heat change in the chemical reaction is:

5606.6J / 0.1 mol = 56066J =

the answer for sodium is 0.186

the answer for rubidium is 0.251

the volume increases