HCl(aq) +NaOH(aq) ---> NaCl +H2O

In a coffee cup calorimeter 50 ml of 2.00M HCl was combined with 50ml of 2.05M NaOH.

a) What is the enthalpy change for the reaction?

b) What is the molar enthalpy for the reaction?

Given:
Change in T= 14 degrees celsius
heat capacity of the calorimeter = 14.0J/degree C
heat capacity of the NaCl solution = 3.91 J/g degree C
Density of 1M Nacl solution = 1.037 g/ml

To find the enthalpy change (ΔH) for the reaction and the molar enthalpy (ΔH/mol) of the reaction, we can use the equation:

ΔH = q / n

where q is the heat transferred in Joules (J), and n is the number of moles of the limiting reactant.

Step 1: Calculate the moles of HCl and NaOH used in the reaction.

For HCl:
Molarity (M) = 2.00 M
Volume (V) = 50 ml = 0.050 L

moles = M x V
moles of HCl = 2.00 M x 0.050 L = 0.100 mol

For NaOH:
Molarity (M) = 2.05 M
Volume (V) = 50 ml = 0.050 L

moles = M x V
moles of NaOH = 2.05 M x 0.050 L = 0.1025 mol

Since the balanced equation shows that they react in a 1:1 ratio, the limiting reactant is HCl with 0.100 mol. This means that the number of moles of NaOH used is also 0.100 mol.

Step 2: Calculate the heat transferred (q) using the equation:

q = CΔT

where C is the heat capacity and ΔT is the change in temperature.

Given:
Change in T = 14 degrees Celsius
Heat capacity of the calorimeter = 14.0 J/degree C

q(calorimeter) = C × ΔT
q(calorimeter) = 14.0 J/degree C × 14 degrees C = 196 J

Step 3: Calculate the mass of the NaCl solution.

Volume of 1M NaCl solution = 50 ml = 0.050 L

mass = density × volume
mass = 1.037 g/ml × 0.050 L = 0.05185 g

Step 4: Calculate the heat capacity of the NaCl solution.

q(NaCl) = mass × C(NaCl)

Given:
Heat capacity of the NaCl solution = 3.91 J/g degree C

q(NaCl) = 0.05185 g × 3.91 J/g degree C × ΔT
q(NaCl) = 0.05185 g × 3.91 J/g degree C × 14 degrees C = 0.9984 J

Step 5: Calculate the total heat transferred (q).

q = q(calorimeter) + q(NaCl)
q = 196 J + 0.9984 J = 196.9984 J

Step 6: Calculate the enthalpy change (ΔH).

ΔH = q / n
ΔH = 196.9984 J / 0.100 mol = 1969.984 J/mol

So, the enthalpy change (ΔH) for the reaction is 1969.984 J/mol.

Step 7: Calculate the molar enthalpy (ΔH/mol).

Since we started with 0.100 mol of HCl and NaOH, the molar enthalpy is the same as the enthalpy change.

Therefore, the molar enthalpy (ΔH/mol) for the reaction is 1969.984 J/mol.