Calculate ΔH for the overall reaction in units of joules

Question

A styrofoam cup calorimeter is used to determine ΔH for the acid-base reaction:

CH₃COOH (aq) + NaOH (aq) ⇔ NaCH₃COO (aq) + H₂O (l)

 

when 25.00mL of 0.500M NaOH at 23.00°C is added to 25.00mL of 0.600M CH₃COOH already in the calorimeter at the same temp, the resulting temp is 25.95°C. The heat capacity of the calorimeter is 27.8J/°C. Assume that the specific heat capacity of the reaction is the same as that of water, 4.184 J/g °C, and that the density of the reaction solution is 1.02g/mL.

 

a. given that the reaction is conducted under constant pressure conditions, calculate ΔH for the overall reaction in units of joules.

 

 

b. calculate ΔH for this reaction in units of kJ/mol NaOH.

Answer 1

Hello Haley

 

For question a), we would need to use this equation, q = c m ΔT where q is the heat released from the reaction, c is the specific heat and ΔT is the change in temperature (T_final - T_initial).

 

ΔT = T_f - T_i = 25.95 - 23.00 = 2.95 ºC

 

The mass of the solution is obtained from the total volume of NaOH and CH₃COOH and the given density.

 

Total volume = V_NaOH + V_CH3COOH = 25.00 + 25.00 = 50.00 mL

 

Since the solution is mostly water, the mass of the solution = density x total volume = 1.02 g/mL x 50.00 mL = 51.00 g.

 

Now let's plug in the numbers into the equation,

 

q = c m ΔT = 4.184 J/gºC x 51.00 g x 2.95 ºC = -629 J.

 

**The sign for q is negative because the reaction is exothermic since the temperature increased.

 

At constant pressure, ΔH = q = -629 J

 

For b), we need to first determine whether NaOH or CH₃COOH is the limiting reagent since the enthalpy value per mole is based on this amount.  For NaOH and CH₃COOH, we are given the molarity and volume.

 

moles = Molarity x Volume (in Liters)

 

NaOH = 0.500 mol/L x 25.00 mL x (1 L / 1000 mL) = 0.0125 mol

 

CH₃COOH = 0.600 mol/L x 25.00 mL x (1 L / 1000 mL) = 0.0150 mol

 

You can see that NaOH is the limiting reagent.  Therefore,

 

ΔH = q / mol = -629 J / 0.0125 mol = -50320 J/mol

 

Since the answer must be in units of kJ/mol, the conversion factor is 1000 J = 1 kJ

 

ΔH = -50320 J/mol x (1 kJ / 1000 J) = -50.32 kJ/mol

 

Hope the above was helpful!